VECTORS AND MOTION IN TWO DIMENSIONS 3

Transcription

1 VECTORS AND MOTION IN TWO DIMENSIONS 3 Q3.. Reason: (a) If one component of the vector is zero, then the other component must not be zero (unless the whole vector is zero). Thus the magnitude of the vector will be the value of the other component. For eample, if A =0mand A = 5 m, then the magnitude of the vector is A = (0 m) + (5 m) = 5 m (b) A zero magnitude sas that the length of the vector is zero, thus each component must be zero. Assess: It stands to reason that a vector can have a nonzero magnitude with one component zero as long as the other one isn t. It also makes sense that for the magnitude of the vector to be zero all the components must be zero. Q3.. Reason: No, it is not possible. A scalar has a magnitude onl but a vector has direction as well. Even if each has the same dimensions, the result of the addition of a scalar and a vector is ambiguous. Assess: We alread learned in chapter that we can t add quantities unless the have the same dimensions; here we also point out that two quantities that will be added (or subtracted) must both be scalars or both vectors. Q3.3. Reason: Consider two vectors A and B. Their sum can be found using the method of algebraic addition. In Question 3. we found that the components of the zero vector are both zero. The components of the resultant of A and B must then be zero also. So R = A + B = 0 R = A + B = 0 Solving for the components of B in terms of A gives B = Aand B = A. Then the magnitude of B is ( B) + ( B) = ( A) + ( A) = ( A) + ( A). So then the magnitude of B is eactl equal to the magnitude of A. Assess: For two vectors to add to zero, the vectors must have eactl the same magnitude and point in opposite directions. Q3.4. Reason: (a) C = A+ B onl if A and B are in the same direction. Size does not matter. (b) C = A B if A and B are in the opposite direction to each other. Size matters onl in that A > B because C as a magnitude can onl be positive. Assess: Visualize the situation with arrows. Q3.5. Reason: The ones that are constant are v, a, and a. Furthermore, a is not onl constant, it is zero. 3-

2 3- Chapter 3 Assess: There are instants when other quantities can be zero, but not throughout the flight. Remember that a = g throughout the flight and that v is constant; that is, projectile motion is nothing more than the combination of two simple kinds of motion: constant horizontal velocit and constant vertical acceleration. Q3.6. Reason: The acceleration of the ball is due to gravit, so the acceleration the ball eperiences is alwas straight downward. (a) The velocit vector of the ball alwas has a component in the horizontal direction since it was thrown at an angle of 40. The horizontal component of the ball s velocit is constant throughout its trajector. Since the velocit vector alwas has a component in the horizontal direction, it is never pointing entirel straight up or down. So there is no point on the trajector where the acceleration and velocit are parallel. Note that if the ball were thrown straight up or down instead of at 40, the ball s velocit and acceleration would be parallel. (b) The acceleration of the ball is alwas straight downward. For the velocit vector to be perpendicular to the acceleration vector, the velocit must point entirel in the horizontal direction. The velocit vector alwas has a component in the horizontal direction, as reasoned in part (a). It has a component in the vertical direction during most of its motion also since the ball travels upward and downward in the vertical direction. There is one point where the ball is not traveling up or down, and that is at the top of its trajector. This is the onl point where the velocit and acceleration are perpendicular. Assess: The acceleration due to gravit alwas points straight downward. See Figure 3.30, which shows the average velocit vectors and acceleration along the trajector of a tossed ball. Q3.7. Reason: B etending their legs forward, the runners increase their time in the air. As ou will learn in chapter 7, the center of mass of a projectile follows a parabolic path. B raising their feet so that their feet are closer to their center of mass, the runners increase the time it takes for their feet to hit the ground. B increasing their time of flight, the increase their range. Also, having their feet ahead of them means that their feet will land ahead of where the would have landed otherwise. Assess: B simpl moving their feet, runners can change their time of flight and change the spot where their feet land. Q3.8. Reason: Running while throwing a ball increases the distance of the throw because it increases the horizontal component of the ball s velocit without changing the time of flight. Relative to the ground, the ball s horizontal velocit component is increased b an amount equal to the speed of the runner. Since the velocit of the runner is purel horizontal, the vertical velocit component of the ball is unchanged and so the time of flight of the ball is unchanged. It is interesting to note that since, relative to the ground, the ball has a greater horizontal component of velocit, the angle of the ball s velocit is higher relative to the person throwing the ball than relative to the ground. However, neither relative to the ground, nor relative to the person throwing the ball will a launch angle of 45 give the maimum range. The angle 45 gives the maimum range for a ball launched b someone not moving because it gives the best compromise between having a high time of flight and a large horizontal component of velocit. But when the person is running, the horizontal component of velocit gets an advantage so the time of flight can be greater at the epense of the horizontal component of velocit. Thus the best angle relative to the person is greater than 45. Assess: While the best launch angle is no longer 45, it is still true that running increases the range of the ball. Q3.9. Reason: The claim is slightl misleading, since the passenger cannot walk at a speed in ecess of 500 mph due to her own efforts relative to the walking surface (the floor of the plane in this case); but she can be walking and moving with such a speed relative to the ground thousands of feet below. Assess: It is important to specif the coordinate sstem when reporting velocities. Q3.0. Reason: The lower the angle of the slope, the lower the acceleration down the slope since acceleration along a ramp is given b g sin θ. Furthermore, the speed at the bottom of a slope will be less if the acceleration is less. Assess: A lower angle gives the skier a lower velocit and, consequentl, better control. Q3.. Reason: The acceleration is due to gravit, so the acceleration will alwas act to pull the cars back down the ramp. Since the roller coaster is constrained to move along the ramp, the acceleration must be along the ramp. So in all three cases the acceleration is downward along the ramp. Assess: Gravit alwas acts, even if motion is constrained b an inclined ramp. See Figure 3.4 in the tet.

3 Vectors and Motion in Two Dimensions 3-3 Q3.. Reason: The time for an object to hit the ground does not depend on its horizontal speed, but onl on its height and initial vertical speed. When the pilot goes twice as fast, all that changes is the horizontal speed of the projectile. Therefore the time of flight will be the same in both cases,.0 s. However the distance travelled horizontall will be doubled since the horizontal speed is doubled and the time of flight is the same. At the doubled speed, the weight will travel twice as far, or 00 m. Assess: The above answer assumed no air resistance. Actuall the greater speed of the faster projectile will slightl increase the time of flight since a faster object eperiences more air resistance. Q3.3. Reason: The cclist s speed ma be constant, but the direction of her motion is alwas changing. Since the direction of her velocit vector is constantl changing, she is alwas accelerating. Assess: Acceleration is caused b an change in velocit, either in magnitude or direction. In circular motion, the acceleration is alwas towards the center of the circle and is called centripetal acceleration. Q3.4. Reason: Because acceleration is defined as the change in the (vector) velocit divided b the corresponding time interval, it, too, is a vector. a = v t A car in uniform circular motion (constant speed) is still accelerating because the direction of the velocit vector is changing. The wa to determine the direction of the acceleration vector is to realize that a must alwas point in the same direction as v because t is not onl a scalar, it is a positive scalar (as long as time doesn t go backward). v = v v, so, to determine the direction of, f i v look at the velocit vector arrows just before and just after the car is pointed north, and subtract them using vector subtraction. v = v v = v + ( v ) and, as shown in the figure, v points west. f i f i Assess: It is worth noting that the specific directions were not as important as the conclusion that the v vector and therefore the a vector both point toward the center of the circle in uniform circular motion. Q3.5. Reason: Since the plane is moving in a circle and constantl changing direction, the plane is constantl accelerating. At an point along its motion, even directl north, the plane is accelerating. For circular motion, the direction of the acceleration is alwas toward the center of the circle. When the plane is headed directl north, the plane s acceleration vector points directl east. See the net figure.

4 3-4 Chapter 3 Assess: In circular motion, there is a centripetal acceleration that alwas points toward the center of the circle. Q3.6. Reason: To make a tighter turn with a smaller radius, ou need to reduce our speed. If ou are traveling at the greatest speed which is safe, then ou are accelerating at the highest acceleration which is safe. Now if the radius is reduced, this tends to increase the acceleration above safe values. To bring it back down, our speed should be reduced. If we solve the formula for centripetal equation, a= v / r, for v, we have: v = ar. So v is proportional to the square root of r. For eample, if we take a turn with a radius which is four times smaller, we need to cut our speed in half. Assess: The maimum safe speed is proportional to the square root of the radius of the turn. Q3.7. Reason: The longest vector will be obtained b turning C around or b turning A around so that the two point in the same direction. The choices A, C, and D all have A and C added together. B is the onl choice in which the two vectors are subtracted and not added together. Since B is perpendicular to A and C, we will get a vector of the same length whether we add or subtract B from the other two. See the figure for the lengths of the different vector combinations. Assess: A longer vector can be created b adding two vectors which point in the same direction. Q3.8. Reason: To generate a vector which points to the left, we could add two vectors which point left, one pointing up and the other down. In C, Q and P fit this description so their sum points to the left. The various vector combinations are shown. Assess: If two vectors have equal and opposite components in a certain direction, sa the direction, then when we add the vectors, the equal and opposite components will cancel and leave us with a vector perpendicular to that direction. Q3.9. Reason: The gas pedal can be considered an accelerator because it can change the velocit of the car. So can an other controls in the car that change the velocit of the car. The brakes certainl can change the velocit b slowing the car.

5 Vectors and Motion in Two Dimensions 3-5 The steering wheel also can change the velocit b changing the direction of travel. The gear shift can also change the speed. (Shifting to a lower gear can slow the car.) So the answer is D, all of the above. Assess: When driving on a straight road at constant speed the gas pedal is not then acting as an accelerator. You have to keep our foot on the gas pedal just to keep the velocit constant, and so, in that scenario, the gas pedal is a nonaccelerator. This question certainl highlights the difference in language between ever da usage and the more specific phsics usage. It pas to know our audience! Q3.0. Reason: The car is traveling at constant speed, so the onl possible cause for accelerations is a change in direction. (a) At point the car is traveling straight to the right on the diagram, so its velocit is straight to the right. The correct choice is B. (b) At point the car is traveling at constant speed and not changing direction so its acceleration is zero. The correct choice is E. (c) The car s velocit at this point on the curve is in the direction of its motion, which is in the direction shown at choice C. (d) The car is moving on a portion of a circle. The acceleration of an object moving in a circle is alwas directl toward the center of the circle. The correct choice is D. (e) The car is moving on a portion of a circle at point 3. The instantaneous velocit vector is directl to the right, which is choice B. (f) The car is accelerating because it is moving on a portion of a circle. The acceleration is toward the center of the circle, which is in direction A. Assess: The instantaneous velocit of a particle is alwas in the direction of its motion at that point in time. For motion in a circle, the direction of the acceleration is alwas toward the center of the circle. Q3.. Reason: (a) The ball is going along the trajector, so the best choice for the velocit at position is D. (b) The direction of the acceleration is not related to the direction of the velocit onl the direction of the change in velocit. The ball is in free fall, and so its acceleration is down, just like the acceleration of all other objects in free fall. The best choice is A. This does mean, however, that the change in velocit vectors ( v ) must also be down. (c) The ball at position 3 is moving horizontall (the vertical component is zero for an instant), so choice C is best. (d) The same argument for the acceleration at position applies at position 3; the acceleration is down. So the answer is A. Assess: Galileo taught us the law of falling bodies: All bodies in free fall (projectiles) have the same acceleration. The ball is in free fall at both position and position 3 so the accelerations are the same, regardless of the fact that the velocities are in different directions. Q3.. Reason: The maimum height the ball reaches onl depends on the initial velocit it had in the vertical direction. The -component of the velocit of this ball is ( v ) = v sin ( θ ) = (3.0 m/s) sin (37.0 ) = 3.8 m/s i i In order to reach the same height when being thrown verticall upward the ball s initial velocit must be 3.8 m/s. The correct choice is A. Assess: Projectile motion is made up of two independent motions: uniform motion at constant velocit in the horizontal direction and free-fall motion in the vertical direction. Q3.3. Reason: The ke to projectile motion problems is to realize that the motion in the -coordinate is independent of the motion in the -coordinate. We can solve an equation in one of these directions and use the results in an equation for the other direction. For eample, t is the same for the horizontal and vertical components of the motion. (a) First find the horizontal component of the velocit, and, realizing it will be constant, find the time to impact. ( v) i = vi cos θ = (89 m/s) cos 40 = 68. m/s distance 300 m time = 4.4 s speed = 68. m/s =

6 3-6 Chapter 3 So the correct choice is C. (b) The vertical component of the initial speed is ( v ) = v sin θ = (89 m/s) sin 40 = 57. m/s i Now that we know t = 4.4s and ( v ) i = 57. m/s we can solve for h (which is f ). v t g t i f = i + ( ) i ( ) = 0.0 m + (57. m/s)(4.4 s) (9.8 m/s )(4.4 s) = 57 m 60 m So the correct choice is D. Assess: The answers to both parts seem reasonable; in either case if we had been off b a factor of 0 in either direction we would think the result not realistic. Q3.4. Reason: The car drops in the vertical direction b a distance of 73 m. Since the car drives off the cliff horizontall, its initial velocit in the vertical direction is 0 m/s. Projectile motion is described b Equations 3.5. (a) To find time given distance we can use f = i + ( v) i t g( t) With i = 0 m, f = 73 m and ( v ) i = 0 m/s. Substituting these values in and solving for t ( 73 m) g (9.80 m/s ) f t = = = 3.9 s The correct choice is C. (b) The horizontal distance traveled b the car is found b multipling v b t : v t = (7 m/s)(3.86 s) = 04 m. Here we have used three figures in t since it is used in an intermediate calculation. To two significant figures, the car lands 00 m from the base of the cliff, so the answer is C. Assess: Projectile motion is made up of two independent motions: uniform motion at constant velocit in the horizontal direction and free-fall motion in the vertical direction. Note that the initial velocit of the car was not relevant to this problem because it was entirel in the horizontal direction. Q3.5. Reason: To the nearest second means we don t even need a calculator if we recall that sin30 =. We can also assume no air resistance. The vertical component of the initial velocit is ( v ) = v sin θ = (0 m/s)( ), so we can analze this as if the projectile had gone straight up with an initial velocit of ( v ) = 0 m/s. i i Understanding that a = g 0 m/s means that the velocit changes b 0 m/s each second. So after s the ball has lost 0 m/s of vertical speed, which makes its vertical speed zero (and consequentl it is at the top of the trajector). We ignore air resistance and sa the motion is smmetric it will take as long to come down as it did to go up, which is one more second. The total is therefore s and the correct choice is B. Assess: The value of g is within about % of 0 m/s, so when we onl want one significant figure we can use that value for ver eas calculations. Even when ou want more significant figures, it is nice to round g to 0 m/s and do a quick mental calculation as an assessment check of our more precise calculator work. Q3.6. Reason: Let us tr to avoid the calculator as in Question 3.5. Since this question asks for range, we need the horizontal component of the ball s speed. For that ou need to know that cos30 = 3 / and that The horizontal component is ( v ) = vi cosθ = ( 0 m/s i )( 3 / ) = 7.3 m/s. Since v is constant (there is no acceleration in the direction), the range of the football is given b ( ) ( )( ) R= v t = 7.3 m/s s = 34.6 m 35 m i Because we have used the time from the previous question, this answer onl has one or two significant figures. The correct choice is C. i

7 Vectors and Motion in Two Dimensions 3-7 Assess: As Questions 3.5 and 3.6 illustrate, b using g 0 m/s, we can estimate the main features of a projectile motion problem such as the time of flight and the range without using a calculator. Q3.7. Reason: The magnitude of the centripetal acceleration is where we want to solve for r: c v ac = r v (4.0 m/s) r = = = 8.0 m a.0 m/s This is r, but the diameter is twice the radius d = 6m, so the correct choice is D. Assess: Think about Ferris wheels ou have been on. One with a radius of 6 m would be reasonable without being overl large. Problems P3.. Prepare: (a) To find A+ B, we place the tail of vector B on the tip of vector A and then connect vector A s tail with vector B s tip. (b) To find A B, we note that A B= A+ ( B). We place the tail of vector B on the tip of vector A and then connect vector A s tail with the tip of vector B. Solve: P3.. Prepare: (a) To find A+ B, we place the tail of vector B on the tip of vector A and connect the tail of vector A with the tip of vector B. Solve: (b) Since A B= A+ ( B), we place the tail of the vector ( B ) on the tip of vector A and then connect the tail of vector A with the tip of vector ( B ). P3.3. Prepare: We can find the positions and velocit and acceleration vectors using a motion diagram.

8 3-8 Chapter 3 Solve: The figure gives several points along the car s path. The velocit vectors are obtained b connecting successive dots. The acceleration vectors are obtained b subtracting successive velocit vectors. The acceleration vectors point toward the center of the diagram. Assess: Notice that the acceleration points toward the center of the turn. As ou will learn in chapter 4, whenever our car accelerates, ou feel like ou are being pushed the opposite wa. This is wh ou feel like ou are being pushed awa from the center of a turn. P3.4. Prepare: Acceleration is found b the method of Tactics Bo 3.. Solve: (a) Let v i be the velocit vector between points 0 and and v f be the velocit vector between points and. (b) Its average speed between points and is probabl greater than its average speed between points 0 and. This is because the distance between and is greater than the distance between 0 and. However, it is possible that the object traveled a greater distance between 0 and than between and if for eample it followed a ver curved path between 0 and. P3.5. Prepare: Acceleration is found b the method of Tactics Bo 3.. Solve: The acceleration vector at each location points directl toward the center of the Ferris wheel s circular motion.

9 Vectors and Motion in Two Dimensions 3-9 Assess: As we will learn later, this acceleration that is directed toward the center is called centripetal acceleration. P3.6. Prepare: The position vector d whose magnitude d is 0 m has an -component of 6 m. It makes an angle θ with the +-ais in the first quadrant. We will use trigonometric relations to find the -component of the position vector. Solve: Using trigonometr, d = d cos θ, or 6 m = (0 m) cos θ. This gives θ = 53.. Thus the -component of the position vector d is d = d sin θ = (0 m) sin 53. = 8 m. Assess: The -component is positive since the position vector is in the first quadrant. P3.7. Prepare: The figure below shows the components v and v, and the angle θ. We will use Tactics Bo 3.3 to find the sign attached to the components of a vector. Solve: We have, v = vsin 40, or 0 m/s = v sin 40, or v = 5.56 m/s. Thus the -component is v = v cos 40 = (5.56 m/s ) cos 40 = m/s. Assess: Note that we had to insert the minus sign manuall with v since the vector is in the fourth quadrant.

10 3-0 Chapter 3 P3.8. Prepare: The figure below shows the components v and v, and the angle θ. We will use Tactics Bo 3.3 to find the sign attached to the components of a vector. Solve: (a) Since v = v cos θ, we have.5 m/s = (3.0 m/s) cos θ θ = cos (.5 m/s 30 m/s) = 33.6 = 34. (b) The vertical component is v = vsin θ = (3.0 m/s) sin 33.6 =.7 m/s. P3.9. Prepare: The figure below shows the components v and v, and the angle θ. We will use Tactics Bo 3.3 to find the sign attached to the components of a vector. Solve: We have v = v + v = v + v. v v cos = θ = (00 m/s) cos 30 = 87 m/s. Assess: For the small angle of 30, the obtained value of 87 m/s for the horizontal component is reasonable. P3.0. Prepare: Vector E points to the left and up, so according to the Tactics Bo 3.3 the components E and E are negative and positive, respectivel. Solve: (a) (b) E = E cos θ and E = E sin θ. E = E sin φ and E = E cos φ. Assess: Note that the role of sine and cosine is reversed because we are using a different angle. θ and φ are complementar angles.

11 Vectors and Motion in Two Dimensions 3- P3.. Prepare: We will follow rules given in the Tactics Bo 3.3. Solve: (a) Vector d points to the right and down, so the components respectivel: d and d are positive and negative, d = d cos θ = (00 m) cos 45 = 70.7 m d = d sin θ = (00 m) sin 45 = 7m (b) Vector v points to the right and up, so the components v and v are both positive: v = v cos θ = (300 m/s) cos 0 = 80 m/s v = v sin θ = (300 m/s) sin 0 = 00 m/s (c) Vector a has the following components: a = a cos θ = (5.0 m/s ) cos 90 = 0 m/s a = a sin θ = (5.0 m/s ) sin 90 = 5.0 m/s Assess: The components have same units as the vectors. Note the minus signs we have manuall inserted according to the Tactics Bo 3.3. P3.. Prepare: We will follow rules given in Tactics Bo 3.. Solve: (a) d = ( km) sin 30 = km d = ( km) cos 30 =.7 km (b) v = (5 cm/s) sin 90 = 5 cm/s v = (5 cm/s) cos 90 = 0 cm/s (c) a = (0 m/s ) sin 40 = 6.4 m/s a = (0 m/s ) cos 40 = 7.7 m/s Assess: The components have the same units as the vectors. Note the minus signs we have manuall inserted according to the Tactics Bo 3.3.

12 3- Chapter 3 P3.3. Prepare: We will draw the vectors to scale as best we can and label the angles from the positive - ais (positive angles go CCW). We also use Equations 3. and 3.. Make sure our calculator is in degree mode. Solve: (a) (b) v v v = ( ) + ( ) = (0 m/s) + (40 m/s) = 45 m/s v 40 m/s θ = tan = tan tan () 63 v = = 0 m/s a= ( a ) + ( a ) = (.0 m/s ) + ( 6.0 m/s ) = 6.3 m/s a 6.0 m/s θ = tan = tan tan ( 3) = 7 = a.0 m/s Assess: In each case the magnitude is longer than either component, as is required for the hpotenuse of a right triangle. The negative angle in part (b) corresponds to a clockwise direction from the positive -ais. P3.4. Prepare: We can use Equations 3. and 3. to find the magnitude and direction of a vector given its components. Solve: (a) See the following diagram.

13 Vectors and Motion in Two Dimensions 3-3 Using Equation 3., v= v + v = + = ( ) ( ) (0 m/s) ( 30 m/s) 3 m/s Using Equation 3., (b) See the following diagram. v 30 m/s θ = tan = tan = 7 v 0 m/s Using Equation 3., Using Equation 3., a= ( a ) + ( a ) = (0 m/s ) + (0 m/s ) = m/s a 0 m/s θ = tan = tan 7 = a 0 m/s Assess: Comparing the vector diagrams to the calculations, both these answers make sense. P3.5. Prepare: Assume ou start at the spot labeled home and that Strawberr Fields and Penn Lane are perpendicular. We will not assume that the lengths in the figure are to scale. We will write each vector in component form for eas addition. We will then need to take the sum and compute the magnitude and direction to report the final answer. The Strawberr Fields vector is S = ( S, S) = (.0 km,0.0 km). The Penn Lane vector is P= ( P, P) = (0.0 km,.0 km). The Abbe Road vector is A= ( A, A) = ( Acos φ, A sin φ) o o = ((4.0 km) sin (40 ),(4.0 km) cos (40 )) = (.57 km, 3.06 km) (The cos and sin are interchanged from Equation 3.0 because φ is measured from the -ais. We have used an etra significant figure for etra accurac.) Solve: Now add the respective components of the three vectors to get the components of the total displacement.

14 3-4 Chapter 3 Now use Equations 3. and 3.3. S = (.0 km, 0.0 km) P = (0.0 km,.0 km) A = (.57 km, 3.06 km) D = (4.57 km,.06 km) D D D = ( ) + ( ) = (4.57 km) + (.06 km) = 5.km D.06 km θ = tan = tan = 4 D 4.57 km where θ is measured ccw from the positive -ais. Assess: Even though the figure ma not be precisel to scale, it, or one ou draw, would convince ou that the answers for the magnitude and direction are both reasonable. P3.6. Prepare: With air resistance and friction ignored, the acceleration down a slope is given b Equation 3.. Assume ou start from rest. Solve: Since ou are moving down the slope, our acceleration along the slope will be The final velocit can be obtained with the equation a= g sin( θ ) = (9.80 m/s ) sin(5 ) =.54 m/s ( v) f = ( v) i + a t = (.54 m/s )(0 s)=5.4 m/s This should be reported as 5 m/s to two significant figures. From Table.3, m/s =.4 mph. Converting to miles per hour, ou are traveling at.4 mph (5.4 m/s) = 57 mph m/s Assess: This answer is reasonable. Compare to Eample 3.7 in the tet. P3.7. Prepare: A visual overview of the car s motion that includes a pictorial representation, a motion diagram, and a list of values is shown below. We have labeled the -ais along the incline. Note that the problem ends at a turning point, where the car has an instantaneous speed of 0 m/s before rolling back down. The rolling back motion is not part of this problem. If we assume the car rolls without friction, then we have motion on a frictionless inclined plane with acceleration a = g sin θ = g sin 5.0 = m/s.

15 Vectors and Motion in Two Dimensions 3-5 Solve: Constant acceleration kinematics gives vi (30 m/s) vf = vi + a( f i) 0 = vi + af f = = = 530 m a ( m/s ) Notice how the two negatives canceled to give a positive value for f. Assess: We must include the minus sign because the a vector points down the slope, which is in the negative -direction. P3.8. Prepare: We can find the acceleration on the ramp and then use the acceleration to find the final velocit of the car. Solve: (a) The maimum possible acceleration is given b the formula a = g sin θ. Plugging in the values of g and θ, we get a =. m/s. (b) The final velocit can be obtained from the formula ( v ) ( v ) = a, using ( v ) = 0 m/s and f i m = 6.8 m, the latter being obtained b converting 55 ft: 55 ft = 55 ft 3.8 ft. m 55 ft = 55 ft The 3.8 ft solution to the first equation is ( v ) = 8.6 m/s. f Assess: As with most ramp problems, this one was best solved b using a rotated coordinate sstem with the -ais along the ramp. P3.9. Prepare: Make a sketch with tilted aes with the -ais parallel to the ramp and the angle of inclination labeled. We must also make a bold assumption that the piano rolls down as if it were an object sliding down with no friction. i As part of the preparation, compute the length of the ramp in the new tilted - coordinates. h.0 m L = = = o.9 m sinθ sin 0

16 3-6 Chapter 3 The acceleration in the new coordinate sstem will be a = g sin θ = (9.8 m/s ) sin 0 = 3.4 m/s. Solve: Since this is a case of constant acceleration we can use the second equation from Table.4 with i = 0.0 m and (v ) i = 0.0 m/s. = a ( t) Solve for t, and use f =.9 m and a = 3.4 m/s, which we obtained previousl. f (.9 m) f t = = = a 3.4 m/s Assess: The ma catch it if the have quick reactions, but the piano will be moving 4.5 m/s when it reaches the bottom. P3.0. Prepare: The acceleration down a ramp is given b Equation 3.. The acceleration of each car depends on the angle of the ramp. The final velocit will depend on the acceleration and the length of the ramp. The cars start from rest, so their initial velocities are all 0 m/s. Solve: The acceleration down a ramp is given b a = g sin ( θ ). The final velocit of the cars can be calculated with Equation.3, ( v) f = ( v) i + a, with ( v ) i = 0 m/s. Solving for ( v), f ( v) f = a. For car A, For car B, For car C, For car D,.3s a= g sin( θ ) = (9.80 m/s ) sin (5 ) =.54 m/s ( v) f = a = (.54 m/s )(0 m) = 7. m/s a= g sin( θ ) = (9.80 m/s ) sin(0 ) = 3.35 m/s ( v) f = a = (3.35 m/s )(0 m) = 8. m/s a= g sin( θ ) = (9.80 m/s ) sin (0 ) = 3.35 m/s ( v) f = a = (3.35 m/s )(8.0 m) = 7.3 m/s a= g sin( θ ) = (9.80 m/s ) sin( ) =.04 m/s ( v) f = a = (.04 m/s )( m) = 7.0 m/s The car with the greatest speed at the bottom of the ramp is car B. Assess: The answer makes sense. Car B and C are both on the ramp with the highest inclination out of all four cars, while car B travels a longer distance than car C. P3.. Prepare: For everda speeds we can use Equation 3. to find relative velocities. We will use a subscript A for Anita and a and a for the respective balls; we also use a subscript G for the ground. We will consider all motion in this problem to be along the -ais (ignore the vertical motion including the fact that the balls also fall under the influence of gravit) and so we drop the subscript. It is also worth noting that interchanging the order of the subscripts merel introduces a negative sign. For eample, v AG = 5 m s, so v GA = 5 m/s. According to Anita means relative to Anita. Solve: For ball : For ball : va = vg + vga = 0 m/s + ( 5 m/s) = 5 m/s va = vg + vga = 0 m/s + ( 5 m/s) = 5 m/s

17 Vectors and Motion in Two Dimensions 3-7 The speed is the magnitude of the velocit, so the speed of ball is 5 m/s. Assess: You can see that at low speeds velocities simpl add or subtract, as the case ma be. Mentall put ourself in Anita s place, and ou will confirm that she sees ball catching up to her at onl 5 m/s while she sees ball speed past her at 5 m/s. P3.. Prepare: We can use the technique of canceling subscripts to find relative velocities. Solve: Anita s friends are standing on the ground, so we can calculate the velocities the threw the balls with b calculating the velocities of the balls relative to the ground. The velocit of ball relative to Anita is (v ) A = +0 m/s. The velocit of ball relative to Anita is (v ) A = 0 m/s. Anita s velocit relative to the ground is (v ) Ag = +5 m/s. Then the velocit of ball relative to the ground is ( v ) = ( v ) + ( v ) =+ 0 m/s + 5 m/s =+ 5 m/s g A Ag The velocit of ball relative to the ground is ( v ) = ( v ) + ( v ) = 0 m/s + 5 m/s = 5 m/s g A Ag Assess: The results make sense. The ball to the left of Anita must be traveling faster than Anita, and the ball to the right must be traveling slower than Anita. P3.3. Prepare: Assume motion along the -direction. The velocit of the boat relative to the ground is (v ) bg ; the velocit of the boat relative to the water is (v ) bw ; and the velocit of the water relative to the ground is (v ) wg. We will use the technique of Equation 3.: ( v ) bg = ( v ) bw + ( v ) wg. Solve: For travel down the river, 30 km ( v) bg = ( v) bw + ( v) wg = = 0.0 km/hr 3.0 hr For travel up the river, 30 km ( v) bg = ( v) bw + ( v) wg = = 6.0 km/hr 5.0 hr Adding these two equations ields ( v ) wg =.0 km/hr. That is, the velocit of the flowing river relative to the earth is.0 km/hr. Assess: Note that the speed of the boat relative to the water downstream and upstream are the same. P3.4. Prepare: We can find the relative velocities using the subscript cancellation formula, tpified b Equation 3.4. Solve: The first drawing in the figure below gives the first part of the race and the second drawing shows the second part. (a) For the first part of the trip, Phillippe s velocit relative to the sidewalk is relative to the floor can be found with subscript cancellation: ( v ) =.0 m/s, his velocit PS

18 3-8 Chapter 3 ( v ) = ( v ) + ( v ) =.0 m/s +.5 m/s = 3.5 m/s Pf PS Sf For the second part of the trip, Phillippe s velocit relative to the sidewalk is ( v ) PS =.0 m/s and his velocit relative to the floor is given b: ( v ) = ( v ) + ( v ) =.0 m/s +.5 m/s = 0.5 m/s Pf PS Sf The time it takes Phillippe to finish the race can be found using the formula t = / v for the two parts of the race: 0 m 0 m tp = + = 46 s 3.5 m/s 0.5 m/s The reason for the negative sign in front of the 0 m in the second fraction is that during the second leg of the trip, Phillippe is going back to his original position. Since his velocit relative to the floor is also negative, the time for that portion of the race comes out positive as it should. Renee s velocit relative to the floor is 0 m/s for the first part of the trip and 0 m/s for the second part. The time it takes Renee to finish the race is as follows: 0 m 0 m tp = + = 0 s.0 m/s.0 m/s Renee wins the race. Her time is 0 s to the end and 0 s back. Her total time is 0 s. Phillippe, on the other hand takes onl 5.7 s to make it to the end but 40 s to make it back, for a total time of 46 s. (b) Renee wins b 6 s. Assess: This problem is reminiscent of the stor of the tortoise and the hare. Renee with her constant speed outperforms Phillippe whose performance was eceptional at first and then less so. P3.5. Prepare: First we can find the velocit of the skdiver with respect to the ground using the idea of cancelling subscripts. Then, knowing the components of this vector, we can find the angle of the vector with the vertical. Solve: (a) The velocit of the skdiver relative to the ground, v SG, can be found since we know his velocit relative to the air: v SA = (0, 5.0 m/s) and the velocit of the air relative to the ground, v AG = (.0 m/s, 0 m/s) : v SG = v SA + v AG = (0 m/s, 5.0 m/s) + (.0 m/s, 0 m/s) = (.0 m/s, 5.0 m/s) The figure shows the three velocit vectors. The skdiver s velocit vector relative to the ground forms the angle φ with the vertical where: v.0 m/s φ tan tan = = = v 5.0 m/s He falls at an angle of to the vertical. (b) The time the skdiver takes to fall is given b

19 Vectors and Motion in Two Dimensions m t = = = 00 s v 5.0 m/s The wind causes the skdiver to drift horizontall at a rate of.0 m/s and he falls for 00 s. B the time he lands, he has drifted through a distance D, which is a product of drift speed and time: D = (.0 m/s)(00 s) = 400 m He will miss his desired landing spot b 400 m. Assess: Notice that the skdiver fell at a constant speed whereas an object in free fall accelerates. This is due to the presence of air resistance. The constant speed at which an object falls when there is air resistance is called terminal speed. P3.6. Prepare: The object is undergoing projectile motion as illustrated in Figure 3.3. Solve: (a) The components of v i are ( v ) = v cos θ = (50 m/s) cos 36.9 = 40.0 m/s i i i ( v ) = v sin θ = (50 m/s) sin 36.9 = 30.0 m/s i Since (v ) i is constant, the position increases b 40.0 m ever second. The -velocit and the -position are obtained from the following equations: ( v ) = ( v ) + a ( t t ) f i f i = + ( v ) ( t t ) + a ( t t ) f i i f i f i Thus, At tf =.0 s: ( v) f = ( v) i+ a( tf ti) = (30.0 m/s) + ( 9.8 m/s )( s 0 s) = 0. m/s At tf =.0 s: f = i+ ( v) i( tf ti) + a ( tf ti) = 0 m + (30.0 m/s)( s 0 s) + ( 9.8 m/s )( s 0 s) = 5. m At tf =.0 s: ( v) f = ( v) i+ a( tf ti) = (30.0 m/s) + ( 9.8 m/s )( s 0 s) = 0.4 m/s At t =.0 s: = + ( v ) ( t t ) + a ( t t ) = 0 + (30.0 m/s)( s 0 s) + ( 9.8 m/s )( s 0 s) = 40.4 m and so on. The table showing,, v, v and v (from t = 0 s to t = 6 s) is given as follows. (b) f f i i f i f i t f t i (s) (m) (m) v (m/s) v (m/s) v (m/s) Assess: As epected the trajector is a parabola. Given an initial velocit of 50 m/s, a value of = 40 m for t = 6 s is reasonable.

20 3-0 Chapter 3 P3.7. Prepare: We will assume the ball is in free fall (i.e., we neglect air resistance). The trajector of a projectile is a parabola because it is a combination of constant horizontal velocit (a = 0.0 m/s ) combined with constant vertical acceleration (a = g). In this case we see onl half of the parabola. The initial speed given is all in the horizontal direction, that is, (v ) i = 5.0 m/s and (v ) i = 0.0 m/s. Solve: (a) (b) (c) (d) This is a two-step problem. We first use the vertical direction to determine the time it takes, then plug that result into the equation for the horizontal direction. = a ( t ) ( 0 m) t = = =.0 s a 9.8 m/s We we use the.0 s in the equation for the horizontal motion. = v t = (5.0 m/s)(.0 s) = 0 m Assess: The answers seem reasonable, and we would get the same answers to two significant figures in a quick mental calculation using g 0m/s. In fact, I did this before computing the algebra so I would know how to scale the graphs. P3.8. Prepare: We can use the vertical part of the motion to calculate the time it takes the ball to hit the floor and the horizontal part of the motion to find out how far from the bench it lands. Solve: Refer to the visual overview shown. The initial vertical velocit is zero. Take the floor as the origin of coordinates. The ball falls from i =.00 m and lands at f = 0 m. (a) We can use the vertical-position equation from Equations 3.5 to find the time it takes the ball to reach the floor. Solving for t, f = i + ( v) i t g( t) 0.00 m =.00 m (9.80 m/s )( t)

21 Vectors and Motion in Two Dimensions 3- (.00m) t = = 0.45s 9.80m/s (b) The distance the ball travels horizontall is governed b the horizontal-position equation from Equations 3.5. = + ( v ) t = (.5 m/s)(0.45 s) = m f i i Assess: This seems reasonable. Compare to Eample 3. in the tet, which is similar. P3.9. Prepare: This problem is asking us for a range so we need the horizontal component of velocit, v, and the time of flight, t. The time of flight, in turn, depends on the initial vertical component of velocit, ( v ) and the overall vertical displacement of the rock,. i Solve: We first find v and ( v ) iusing Equations 3.5: v = vi cos θ = (5 m/s) cos 30 =.7 m/s (v ) = v sin θ = (5 m/s) sin 30 =.5 m/s i i Using the equation which relates vertical velocit, acceleration and displacement: can find the final vertical component of velocit. v = ( ) f (.5 m/s) ( 9.8 m/s )( m) v v = a, we ( ) f ( ) i This equation has two solutions: ( v ) = ± 9.8 m/s. We choose the negative solution since the rock is descending when it lands. Now we can find the time of flight using a = v / t: Finall, the range of the rock is given b f t = v a = = / ( 9.8 m/s.5 m/s) / ( 9.8 m/s ) 3.9 s = v t: = v t = (.7 m/s)(3.9 s) = 7 m The rock lands 7 m from the castle wall. Assess: Notice that the mass of the rock was not involved. So if this had been a baseball thrown from a m tower, it would have gone equall far about three quarters of a foot ball field.

22 3- Chapter 3 P3.30. Prepare: We will appl the constant-acceleration kinematic equations to the horizontal and vertical motions as described b Equations 3.5. Solve: Using = + v t t + a t t we get f A ia ( ) i A( fa ia) ( ) A( fa ia),.0 m = 0 m + 0 m + ( 9.8 m/s )( t 0s) tfa = 0.45 s = tf B fa Because fa = fb, so both take the same time to reach the floor. We are now able to calculate fa and fb as follows: fa = ia + ( ) i A( fa ia) + ( ) A( fa ia) = 0 m + (5.0 m/s)(0.45 s 0 s) + 0 m =.3 m v t t a t t = + ( v ) ( t t ) + ( a ) ( t t ) = 0 m + (.5 m/s)(0.45 s 0 s) + 0 m =. m fb ib i B fb ib B fb ib Assess: Note that t fb = t fa since both the spheres move with the same vertical acceleration and both of them start with zero vertical velocit. The horizontal distance for sphere B is one-half the distance for sphere A because the horizontal velocit of sphere B is one-half that of A. P3.3. Prepare: We will appl the constant-acceleration kinematic equations to the horizontal and vertical motions as described b Equations 3.5. The effect of air resistance on the motion of the bullet is neglected. Solve: (a) Using (b) Using = + v t t + a t t we obtain f i ( ) i( f i) ( f i), (.0 0 m) = 0 m + 0 m + ( 9.8 m/s )( t 0 s) tf = s v t t a t t f = i + ( ) i( f i) + ( f i), f (50 m) 0 m ( ) ( s 0 s) 0 m ( ) = 78 m/s = + v i + v i Assess: The bullet falls cm during a horizontal displacement of 50 m. This implies a large initial velocit, and a value of 78 m/s is not surprising.

23 Vectors and Motion in Two Dimensions 3-3 P3.3. Prepare: We are asked to find the take-off speed and horizontal speed of the kangaroo given its initial angle, 0, and its range. Since the horizontal speed is given b v = vcosθ and the time of flight is given b t = vsin θ / g, the range of the kangaroo is given b the product of these: = vsin θcos θ / g. Solve: (a) We can solve the above formula for v and then plug in the range and angle to find the take-off speed: ( ) ( ) v= g / sin θ cos θ = 9.8 m/s (0 m)/( sin 0 cos 0 ) =.3 m/s Its take-off speed is m/s, to two significant figures. (b) Its horizontal speed is given b v = v cos θ = (.3 m/s) cos 0 =.6 m/s or m/s to two significant figures. Assess: The reason the horizontal speed and take-off speed appear the same is that 0 is a small angle and the cosine of a small angle is approimatel equal to. P3.33. Prepare: The golf ball is a particle following projectile motion. We will appl the constantacceleration kinematic equations to the horizontal and vertical motions as described b Equations 3.5. Solve: (a) The distance traveled is f = (v i ) t f = v i cos θ t f. The flight time is found from the -equation, using the fact that the ball starts and ends at = 0: Thus the distance traveled is For θ = 30, the distances are The flight times are vi sin θ f i = 0 = vi sin θ tf gt f = ( vi sin θ gt f) tf tf = g v cos θ i f = i = i f earth gearth 9.80 m/s v sin θ vi sin θ cos θ g g v sin θ cos θ (5 m/s) sin 30 cos 30 ( ) = = = 55. m v sinθ cosθ v sinθ cosθ v sinθ cosθ ( ) = = = 6 = 6( ) = 33. m i i i f moon f earth gmoon g 6 earth gearth v sin θ i ( tf ) earth = =.55 s gearth v sin θ v sin θ ( t ) = = = 6( t ) = 5.30 s i i f moon f earth gmoon 6 gearth The ball spends 5.30 s.55 s =.75 s = 3 s longer in flight on the moon. (b) From part (a), the distance traveled on the moon is 33 m or 330 m to two significant figures. (c) From part (a), the golf ball travels 33. m 55. m = 76 m farther on the moon than on earth. P3.34. Prepare: We can use Equation 3.6. Solve: (a) Converting revolutions per minute to revolutions per second revolutions minute rev/s 3 minute = 60 s

24 3-4 Chapter 3 (b) Using Equation 3.6 T =.8 s f = 0.56 rev/s = Assess: This seems reasonable, if ou re old enough to remember LPs. P3.35. Prepare: We need to convert the 5400 rpm to different units and then find the period which is the inverse of frequenc. Solve: (a) The hard disk s frequenc can be converted as follows: rev rev min rev = 90 min min = 60 sec sec Its frequenc is 90 rev/s. (b) Rewriting Equation 3.6, we have the following: T = ms f = 90 rev/s = Its period is ms. Assess: This is about the rate that the engine in a car turns if it is straining. So an automobile engine completes a ccle ever 0 or 0 ms. P3.36. Prepare: We can use the formula for centripetal acceleration, Equation We need the radius of the track which is half the diameter: R= D/ = (45 m)/ =.5 m. Solve: v (5 m/s) a = = = 0 m/s r.5 m We now convert this acceleration to units of g: g = = 9.8 m/s 0 m/s 0 m/s.0 Assess: The grehounds are accelerating at around the acceleration of free fall! P3.37. Prepare: We are asked to find period, speed and acceleration. Period and frequenc are inverses according to Equation 3.6. To find speed we need to know the distance traveled b the speck in one period. Then the acceleration is given b Equation Solve: (a) The disk s frequenc can be converted as follows: The period is the inverse of the frequenc: rev rev min rev 0,000 0,000 = 67 min min = 60 sec sec T = 6.00 ms f = 67 rev/s = (b) The speed of the speck equals the circumference of its orbit divided b the period: which rounds to 63 m/s. πr π(6.0 cm) 000 ms m v = = 6.8 m/s, T 6.00 ms s = 00 cm (c) From Equation 3.30, the acceleration of the speck is given b v / r : v (6.8 m/s) 00 cm a = = r 6.0 cm = m g 65,700 m/s,

25 Vectors and Motion in Two Dimensions 3-5 Which rounds to 66,000 m/s. In units of g, this is as follows: g = = 9.8 m/s 65,700 m/s 65,700 m/s 6,700 Assess: The speed and acceleration of the edge of a CD are remarkable. The speed, 63 m/s, is about 40 mi/hr. As ou will learn in chapter 4, ver large forces are necessar to create large accelerations like 6,700 g. g P3.38. Prepare: The acceleration the rider eperiences in the centrifuge is a centripetal acceleration. Solve: Since a = v /r, we have v ar v = = (98 m/s )( m) = 34 m/s Assess: 34 m/s 70 mph is a large et understandable speed. P3.39. Prepare: Eamine the formula carefull. a = v r Before plugging in numbers, notice that if the speed is held constant (as in part (a)), then a and r are inversel proportional to each other: doubling one halves the other. And if r is held constant (as in part (b)), then there is a square relationship between a and v: doubling v quadruples a. Solve: It is convenient to use ratios to solve this problem, because we never have to know an specific values for r or v. We ll use unprimed variables for the original case (a = 8.0 m/s ), and primed variables for the new cases. (a) With the speed held constant, v = v but r = r. So a = a= (8.0 m/s ) = 4.0 m/s. (b) With the radius held constant, r = r but v = v. a a a a = = = v v r r v v r r = = = = v ( v) r r v v r r 4 So a = 4a= 4(8.0 m/s ) = 3 m/s. Assess: Please familiarize ourself with this ratio technique and look for opportunities to use it. Again, the advantage is not needing to know an specific values of r or v not onl not having to know them, but realizing that the result is independent of them. The dail life lesson is that driving around a curve with a larger radius produces gentler acceleration. Going around a given curve faster, however, requires a much larger acceleration (produced b the friction between the tires and the road), and the relationship is squared. If our tires are bald or the road slipper there won t be enough friction to keep ou on the road if ou go too fast. And remember that it is a squared relationship, so going around a curve twice as fast requires four times as much friction. P3.40. Prepare: From Equation 3.30, centripetal acceleration is directl proportional to velocit squared. Solve: Since a is proportional to v, if v is doubled, a is quadrupled to m/s. One wa to see this is to look at the ratio of the velocit after the doubling to the velocit before the doubling and also the acceleration after the doubling to the acceleration before the doubling: a v / r v v = = = a v / r v v This tells us that if we increase v b an factor, a will be increased b the square of that factor. Assess: This shows how important it is to avoid making turns too fast. Centripetal acceleration depends ver sensitivel on speed. As ou will learn in chapter 5, the centripetal acceleration of a car is provided b friction and if the friction cannot provide the needed acceleration, m/s in this case, the car will run off the road.

26 3-6 Chapter 3 P3.4. Prepare: The magnitude of centripetal acceleration is given in Equation Solve: The centripetal acceleration is given as.5 times the acceleration of gravit, so Using Equation 3.30, the radius of the turn is given b Assess: This seems reasonable. a = (.5)(9.80 m/s ) = 5 m/s v (0 m/s) r = = = 7 m a 5 m/s P3.4. Prepare: The vectors AB,, and C = A+ B are shown. Because A= A + A and B= B + B, so the components of the resultant vector are C = A + Band C = A + B. Solve: (a) With A = 5, A =, B = 3, and B = 5, we have C = and C = 3. (b) Vectors A, B, and C are shown with their tails together. (c) Since C =, and C = 3, the magnitude and direction of C are C 3 θ = tan = tan 56 C = C = () + ( 3) = 3.6. Assess: The vector C is to the right and down, thus impling a negative -component and positive -component, as obtained above. Alsoθ > 45 since C > C. P3.43. Prepare: The vectors AB,, and D= A B are shown. Because A= A + A and B= B + B, the components of the resultant vector are D = A B and D = A B. so

27 Vectors and Motion in Two Dimensions 3-7 Solve: (a) With A = 5, A =, B = 3, and B = 5, we have D = 8 and D = 7. (b) Vectors A, B and D are shown in the above figure. (c) Since D = 8 and D = 7, the magnitude and direction of D are D 7 D = (8) + (7) = θ = tan = tan = 4 D 8 Assess: Since D < D, the angle θ is less than 45, as it should be. P3.44. Prepare: Because A= A + A, and B= B + B, so the components of the resultant vector are E = A + 3B and E = A + 3B. The vectors AB,, and E = A+ 3B are shown. Solve: (a) With A = 5, A =, B = 3, and B = 5, we have E = and E =. (b) Vectors A, B, and E are shown in the above figure. (c) From the E vector, E = and E =. Therefore, the magnitude and direction of E are E = () + ( ) = E φ = tan = tan = 5. E Assess: Note that φ is the angle made with the -ais, and that is whφ = tan ( E / E ) rather than tan ( E / E) which would be the case if φ were the angle made with the -ais. P3.45. Prepare: Refer to Figure P3.45 in our tetbook. Because A= A + A, B= B + B, and C = C + C, so the components of the resultant vector are D = A + B + C and D = A + B + C. D and D are given and we will read the components of A and C off Figure P3.45. Solve: (a) A = 4, C = 0, and D =, so B = A C + D =. Similarl, A = 0, C =, and D = 0, so B = A C + D =.

28 3-8 Chapter 3 (b) With the components in (a), B = ( ) + () =.8 B θ = tan = tan = 45 B Since B has a negative -component and a positive -component, the angle θ made b B is with the -ais and it is above the -ais. Assess: Since B = B, θ = 45 as is obtained above. P3.46. Prepare: The vectors AB,, and C are shown. We will first calculate the - and -components of each vector and then obtain the magnitude and the direction of the vector D. Solve: (a) The vectors A, B, and C are drawn above. (b) The components of the vectors A, B, and C are A = (3 m) cos 0 =.8 m and A = (3 m) sin 0 =.03 m; B = 0m and B = m; C = (5 m) cos 70 =.7m and C = (5 m) sin 70 = 4.70 m. (c) We have D= A+ B+ C = D + D, which means D =. and D = D = (.m) + (3.73 m) = 3.9 m θ = tan = tan 3.36 = 73. The direction of D is south of east, 73 below the positive -ais. P3.47. Prepare: We need to draw a right triangle which represents the staircase. The hpotenuse is given b multipling speed b distance and then the other two sides are given from trigonometric ratios. Solve: The distance traveled, D, is speed multiplied b time: D = (3.5 m/s)(.0 s)=7.0 m. This is the hpotenuse of the staircase. The height gained is opposite the 38 angle and so is obtained using the sine function: = (7.0 m) sin 38 = 4.3 m The horizontal distance travelled is adjacent to the 38 angle, so we need to use the cosine function.

29 Vectors and Motion in Two Dimensions 3-9 = (7.0 m/s) cos 38 = 5.5 m Assess: It makes sense that the horizontal distance traveled is greater than the height gained since the angle of the stairs, 38, is less than 45. At that angle, the vertical and horizontal distances would have been equal. P3.48. Prepare: The minute hand of the watch is shown in the figure. Solve: (a) We have S 8:00 = (.0 cm, north) displacement vector is and S 8:0 = (.0 cm cos 30, east) + (.0 cm sin 30, south). The r = S8:0 S8:00 = (.74 cm, east) + (3.00 cm, south) (b) We have S 8:00 = (.0 cm, north) and S 9:00 = (.0 cm, north). The displacement vector is r = S9:00 S8:00 = 0. Assess: The displacement vector in part (a) has positive -component (toward east) and negative -component (toward south). The vector thus is to the right and points down, in the IV quadrant. This is what the vector drawn from the tip of the 8:00 A.M. arm to the tip of the 8:0 A.M. arm will be. P3.49. Prepare: In the coordinate sstem shown below, the mouse starts from the origin, so his initial position vector is zero. His net displacement is then just the final position vector, which is just the sum of the three vectors A, B, and C. Solve: We are given A = (5 m, east) and C = ( m, down). Using trigonometr, B = (3 cos 45 m, east) + (3 sin 45 m, south). The total displacement is r = A+ B+ C = (7. m, east) + (. m, south) + (m, down). The magnitude of r is r = (7.) + (.) + () m = 7.5 m Assess: A displacement of 7.5 m is a reasonable displacement.

30 3-30 Chapter 3 P3.50. Prepare: The ideas of vector addition and subtraction can be used here to find the pilot s third displacement. Solve: See the following diagram. We place the origin of coordinates at the origin of the plane s trip. Note that northeast means eactl 45 north of east. The three legs of his trip are labeled D, D, and D 3. His net displacement is labeled b R. His trip and net displacement can be described b the vector equation R= D + D + D 3 We need to calculate D. 3 Solving the equation for D 3 we find D = R D D 3 We can use the technique of algebraic addition to solve this equation. The components of each of the vectors on the right hand side of the equation are R = 0.00 km R = 70.0 km D = Dcos ( θ ) = (6.0 km) cos (45 ) = 8.4 km D = Dsin ( θ ) = (6.0 km) sin (45 ) = 8.4 km D = 0.00 km D = 45.0 km We can calculate the components of D 3 b subtracting the components of D and D from the components of R. The vector is drawn as shown. D3 = R D D = 0.00 km 8.4 km 0.00 km = 8.4 km D = R D D = 70.0 km 8.4 km 45.0 km = 6.6 km 3

31 Vectors and Motion in Two Dimensions 3-3 The magnitude of 3 D is ( ) ( ) D3 = D3 + D3 = 9.5 km. From the previous diagram, the direction of D 3 is θ = D = 6.6 km = 3 tan tan 0 north of west D3 8.4 km Assess: This is reasonable. The pilot must travel west to arrive directl north of his starting point, and must have continued north for a few kilometers to be so far north of his starting point. P3.5. Prepare: We draw a picture of the plane s path. The distance traveled b the plane is given b speed multiplied b time and forms the hpotenuse. The shortest distance to the equator is 00 km and this is the length of the side adjacent to the 30 angle. We can relate these two sides of a triangle using trigonometr. Solve: If D is the distance traveled b the airplane, then we can write cos30 = (00 km)/d and solve for the distance traveled: D = (00 km)/cos 30 = 5 km The time of flight is equal to the distance traveled divided b speed: Finall, we convert this time to minutes: It takes the pilot 46 min to reach the equator. t = (5 km)/(50 km/hr) = hr 60 min hr = hr = 46 min hr

32 3-3 Chapter 3 Assess: The pilot s time to reach the equator is greater than it would be if he were fling directl toward the equator. If the flight were direct, the distance would be 00 km instead of 5 km and the time of flight would be 40 min instead of 46 min. P3.5. Prepare: We will need to add the individual displacements as vectors to calculate the bacterium s net displacement. Solve: (a) The components of the displacement vectors can be read directl from Figure P3.5. D =+ 50 µ m D =+ 0 µ m AB AB D = 0µ m D = + 0µ m BC BC D =+ 40 µ m D =+ 0 µ m CD CD D = 50 µ m D = 50 µ m DE Note both of the components of D 4 are negative. Note also that the components are read from the tail to the tip of the vector relative to the tail. The bacterium is moving with constant speed v = 0 µ m/s along each portion of the path. We will calculate the angles in the following figure in order to calculate components of the velocit. During each segment of the motion, the bacterium is headed eactl in the direction of its displacement. DE For D, the angleθ AB AB vector v are AB The displacement D BC indicated isθ = tan (0 µ m/50 µ m) = so the components of the velocit v v AB AB For D, the angle θ CD CD indicated is vector v CD are AB = v cos ( θ ) = (0 µ m/s) cos ( ) = 9.6 µ m/s AB = vsin ( θ ) = ( µ m/s) sin ( ) = 3.8 µ m/s AB has no -component, so the components of velocit vector v CD are v = 0.00 µ m/s BC v = 0.0 µ m/s CD BC θ = µ µ so the components of the velocit tan (0 m/40 m) =4 v = v cos ( θ ) = (0 µ m/s) cos (4 ) = 9.4 µ m/s CD CD v = vsin ( θ ) = (0 µ m/s) sin (4 ) = 4.84 µ m/s CD CD For D, the angle θ DE DE indicated is θ = tan (50 µ m/50 µ m)=45 so the components of the velocit DE vector v DE are

33 Vectors and Motion in Two Dimensions 3-33 v v DE DE = v cos ( θ ) = (0 µ m/s) cos (45 ) = 4.µ m/s DE = vsin ( θ ) = (0 µ m/s) sin (45 ) = 4.µ m/s DE Additional significant figures have been kept in reporting the components. Each should be reported to two significant figures. (b) We need to calculate the length of each displacement vector using the components obtained in part (a). ( ) ( ) AB AB AB BC = 0 µ m ( ) ( ) CD CD CD ( ) ( ) DE DE DE D = D + D = + = D (50 µ m) (0 µ m) 5.0 µ m D = D + D = µ + µ = µ (40 m) (0 m) 4. m D = D + D = µ + µ = µ ( 50 m) ( 50 m) 70.7 m Additional significant figures have been reported in these results for use later. Each of these distances should be reported to two significant figures. The total distance traveled is the sum of these distances, 5.0 µ m + 0 µ m + 4. µ m µ m=7.9 µ m, which should be reported as70 µ m to two significant figures. We will use the more precise result in part (c). The magnitude of the net displacement can be found from the components of the vector sum of the four vectors in the diagram. The components of the resultant D are the sum of the components of the displacements found in part (a). D = D + D + D + D =+ 50 µ m µ m + 40 µ m 50 µ m = 40 µ m AB BC CD DE D = D + D + D + D =+ 0 µ m + 0 µ m + 0 µ m 50 µ m = 0 µ m AB BC CD DE The components of the resultant could have also been read directl from the diagram. The magnitude of the resultant is D= D + D = µ + µ = µ ( ) ( ) (40 m) ( 0 m) 44.7 m Where additional figures have been kept for use in part (c). This should be reported to two significant figures as D = 45 µ m. (c) Since the bacterium s speed is constant, its average speed is 0 µ m. The bacterium s average velocit would be the magnitude of its net displacement divided b the duration of its travel. The total time it took to travel is given in terms of the total distance it traveled divided b its speed. 7.9 µ m t = = 8.6 s 0 µ m/s Its net displacement was calculated as 44.7 µ m so its average velocit is D 44.7 µ m v = = = 5. µ m/s t 8.6 s From Equation 3., the average velocit is in same direction as the net displacement. Refer to the diagram above. The angle in the diagram is given b D 0 µ m θ = tan = tan = 7 D 40 µ m Assess: The answers here make sense. In part (c), the average velocit is much less than the speed of the bacterium because while the bacterium has traveled a large distance, it s displacement from the origin is small. Note the direction of the velocit vector is the same as the direction of the net displacement. P3.53. Prepare: The skier s motion on the horizontal, frictionless snow is not of an interest to us. The skier s speed increases down the incline due to acceleration parallel to the incline, which is equal to g sin0. A visual overview of the skier s motion that includes a pictorial representation, a motion representation, and a list of values is shown.

34 3-34 Chapter 3 Solve: Using the following constant-acceleration kinematic equations, v = v + a ( ) f i f i (5 m/s) (3.0 m/s) (9.8 m/s ) sin 0 ( f 0 m) f 64 m = + = v = v + a ( t t ) f i f i (5 m/s) (3.0 m/s) (9.8 m/s )(sin 0 ) tf tf 7. s = + = Assess: A time of 7. s to cover 64 m is a reasonable value. P3.54. Prepare: A visual overview of the skier s motion that includes a pictorial representation, a motion diagram, and a list of values is shown. We can view this problem as two one-dimensional motion problems. The horizontal segments do not affect the motion because the speed does not change. So, the problem starts at the bottom of the uphill ramp and ends at the bottom of the downhill ramp. The final speed from the uphill roll (first problem) becomes the initial speed of the downhill roll (second problem). Because the aes point in different directions, we can avoid possible confusion b calling the downhill ais the z-ais and the downhill velocities u. The uphill ais as usual will be denoted b and the uphill velocities as v. Note that the height information, h = m, has to be transformed into information about positions along the two aes. Solve: (a) The uphill roll has a0 = g sin α = 6.93 m/s. The speed at the top is found from v = v + a ( ) v = v + a = (5 m/s) + ( 6.93 m/s )(.44 m) =.3 m/s.3 m/s (b) The downward roll starts with velocit u = v =.3 m/s and a =+ g sin β = 4.90 m/s. Then, u = u + a ( z z ) = (.3 m/s) + (4.90 m/s )(.00 m 0 m) u = 5.0 m/s (c) The final speed is equal to the initial speed, so the percentage change is zero! This result ma seem surprising, but can be more easil understood after we introduce the concept of energ. P3.55. Prepare: A visual overview of the puck s motion that includes a pictorial representation, a motion diagram, and a list of values is shown as follows.

35 Vectors and Motion in Two Dimensions 3-35 Solve: The acceleration, being the same along the incline, can be found as v = v + a ( ) (4.0 m/s) = (5.0 m/s) + a(3.0 m 0 m) a=.5 m/s 0 0 We can also find the total time the puck takes to come to a halt as v = v + at ( t) 0 m/s = (5.0 m/s) + (.5 m/s ) t t = 3.3 s 0 0 Using the above obtained values of a and t, we can find as follows: = 0 + v0( t t0) + a( t t0) = 0 m + (5.0 m/s)(3.3s) + (.5 m/s )(3.3s) = 8.3 m That is, the puck goes through a displacement of 8.3 m. Since the end of the ramp is 8.5 m from the starting position 0 and the puck stops 0. m or 0 cm before the ramp ends, ou are not a winner. P3.56. Prepare: Assume motion along the -direction. Let f i be the displacement from our gate to the baggage claim. We will use the technique of Equation 3.: ( v ) g = ( v ) m + ( v ) mg. Solve: In the first case, when the moving sidewalk is broken, we can find our velocit v Y ( f i) = 50 s In the second case, when ou stand on the moving sidewalk, the velocit of the sidewalk relative to the ground is f i vsg = 75 s In the third case, when ou walk while riding, we can use the equation () v = () v + () v g s sg That is, our velocit relative to the ground, when ou are walking on the moving sidewalk, is equal to our velocit relative to the moving sidewalk (which is v Y ) plus the sidewalk s velocit relative to the ground. Thus, Assess: A time smaller than 50 s was epected = + t = 30 s t 50 s 75 s P3.57. Prepare: Both ships have a common origin at t = 0 s. Solve: (a) The velocit vectors of the two ships are as follows: va = (0 mph cos 30, north) + (0 mph sin 30, west) = (7.3 mph, north) + (0.0 mph, west) vb = (5 mph cos 0, north) + (5 mph sin 0, east) = (3.49 mph, north) + (8.55 mph, east) Since r = v t, r = v A A ( hr) = (34.64 mph, north) + (0.0 mph, west) r = v ( hr) = (46.98 miles, north) + (7.0 miles, east) B B So, the displacement between the two ships is R= ra rb = (.34 miles, south) + (37.0 miles, west) R= 39.miles

36 3-36 Chapter 3 (b) The speed of A as seen b B is: V = va vb = (6.7 mph, south) + (8.55 mph, west) V = 9.5 mph. Assess: The value of the speed is reasonable. P3.58. Prepare: The boat s motion is shown below. Solve: (a) The river is 00 m wide. If Mar rows due north at a constant speed of.0 m/s, it will take her 50 s to row across. But while she s doing so, the current sweeps her boat sidewas a distance m/s 50 s = 50 m. Mar s net displacement is the vector sum of the displacement due to her rowing plus the displacement due to the river s current. She lands 50 m east of the point that was directl across the river from her when she started. (b) Mar s net displacement is shown on the figure. P3.59. Prepare: A visual overview of the ducks motion is shown below. The resulting velocit is given b v = v + v, fl wind where v wind = 6 m/s, east) and vfl = ( vfl sin θ, west) + ( vfl cos θ, south). Solve: Substituting the known values we get v = (8 m/s sin θ, west) + (8 m/s cos θ, south) + (6 m/s, east). That is, v = ( 8 m/s sin θ, east) + (8 m/s cos θ, south) + (6 m/s, east). We need to have v = 0. This means 0 = 8 m/s 6 sin θ + 6 m/s, so sinθ = andθ = Thus the ducks should head 48.6 west of south. 8 P3.60. Prepare: The kaaker s speed of 3.0 m/s is relative to the water. Since he s being swept toward the east, he needs to point at angle θ west of north. The direction the kaaker should paddle can be obtained from the technique of Equation 3.: v = v + v, kg kw wg where vkg = ( vkg, north) and vwg = ( vwg, east) with v wg =.0 m/s. Also, v kw = [( vkw cos θ,north) + ( vkw sin θ,west)] with v kw = 3.0 m/s. Solve: (a) v = v + v, kg kw wg ( v kg, north) = [(3.0 m/s cos θ, north) + (3.0 m/s sin θ, west)] + (.0 m/s, east) = [( 3.0 sin θ +.0) m/s, east] + (3.0 cos θ m/s, north) In order to go straight north in the earth frame, the kaaker s velocit along the east must be zero. This will be true if.0.0 θ = θ = 3.0 = = 3.0 sin sin Thus he must paddle in a direction 4 west of north. (b) His northward speed is v kg = 3.0 cos (4.8 ) m/s =.36 m/s. The time to cross is