2.1. Model: The car is represented by the particle model as a dot.

Transcription

1 .. Model: The car is represented by the particle model as a dot. Solve: (a) Time t (s) Position x (m) (b)

2 .. Solve: Diagram Position Velocity Acceleration (a) (b) (c) Negative Negative Positive Positive Negative Negative Positive Negative Negative

3 .3. Solve: A forgetful physics professor goes for a walk on a straight country road. Walking at a constant speed, he covers a distance of 300 m in 300 s. He then stops and watches the sunset for 00 s. Finding that it was getting dark, he walks faster back to his house covering the same distance in 00 s.

4 .4. Solve: Forty miles into a car trip north from his home in El Dorado, an absent-minded English professor stopped at a rest area one Saturday. After staying there for one hour, he headed back home thinking that he was supposed to go on this trip on Sunday. Absent-mindedly he missed his exit and stopped after one hour of driving at another rest area 0 miles south of El Dorado. After waiting there for one hour, he drove back very slowly, confused and tired as he was, and reached El Dorado in two hours.

5 .5. Model: We will consider the car to be a particle that occupies a single point in space. Visualize: Solve: Since the velocity is constant, we have x = x + v t. Using the above values, we get f i x = 0 m + ( 0 m/s)( 45s) = 450 m Assess: 0 m/s mph and implies a speed of 0.4 miles per minute. A displacement of 450 m in 45 s is reasonable and expected. x

6 .6. Model: We will consider Larry to be a particle. Visualize: Solve: Since Larry s speed is constant, we can use the following equation to calculate the velocities: v s sf si = t t (a) For the interval from the house to the lamppost: 00 yd 600 yd v = = 00 yd/ min 907 : 905 : For the interval from the lamppost to the tree: 00 yd 00 yd v = =+ 333 yd/ min 90 : 907 : (b) For the average velocity for the entire run: 00 yd 600 yd v avg = =+ 0 yd/ min 90 : 905 : f i

7 .7. Model: Cars will be treated by the particle model. Visualize: Solve: Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows: x x x0 x x0 v = = t = t0 + t t t0 v Using the known values identified in the pictorial representation, we find: xalan xalan mile talan = talan 0 + = 8:00 AM + = 800 : AM + 8 hr = 4:00 PM v 50 miles/hour xbeth xbeth mile tbeth = tbeth 0 + = 9:00 AM + = 9:00 AM hr = 3:40 PM v 60 miles/hour (a) Beth arrives first. (b) Beth has to wait t Alan t Beth = 0 minutes for Alan. Assess: Times of the order of 7 or 8 hours are reasonable in the present problem.

8 .8. Model: The bicyclist is a particle. Visualize: Please refer to Figure Ex.8. Solve: The slope of the position-versus-time graph at every point gives the velocity at that point. The slope at t = 0 s is s 00 m 50 m v = = =.5 m/s t 0 s The slope at t = 5 s is 00 m 00 m v = = 0 m/s 0 s The slope at t = 35 s is 0 m 00 m v = = 0 m/s 0 s

9 .9. Solve: (a) The time for each of segment is t = 50 mi/40 mph = 5/4 hr and t = 50 mi/60 mph = 56 / hr. The average speed to the house is 00 mi = 48 mph 5/6 hr + 5/4 hr (b) Julie drives the distance x in time t at 40 mph. She then drives the distance x in time t at 60 mph. She spends the same amount of time at each speed, thus t = t x/40 mph = x/60 mph x = ( 3 / ) x But x + x = 00 miles, so (/3) x + x = 00 miles. This means x = 60 miles and x = 40 miles. Thus, the times spent at each speed are t = 40 mi/40 mph =.00 hr and t = 60 mi/60 mph =.00 hr. The total time for her return trip is t + t =.00 hr. So, her average speed is 00 mi/ hr = 50 mph.

10 .0. Visualize: Please refer to Figure Ex.0. Solve: (a) We can obtain the values for the velocity-versus-time graph from the equation v = s/ t. (b) There is only one turning point. At t = s the velocity changes from +0 m/s to 0 m/s, thus reversing the direction of motion. At t =, there is an abrupt change in motion from rest to +0 m/s, but there is no reversal in motion.

11 .. Visualize: Please refer to Figure Ex.. The particle starts at x 0 = 0 m at t 0 = 0. Its velocity decreases as time increases, goes through zero at t = 3 s, and then reverses direction. Solve: (a) Using the equation x f = x 0 + area of the velocity graph between t i and t f, x s = 0 m + area of trapazoid between 0 s and s = 0 m + ( m/s + 4 m/s )( s ) = 6 m x3 s = 0 m + area of triangle between 0 s and 3 s = 0 m + ( m/s )( 3 s ) = 8 m x4 s = x3 s + area between 3 s and 4 s = 8 m + ( 4 m/s )( s ) = 6 m (b) The particle reverses direction at t = 3 s.

12 .. Visualize: Please refer to Figure Ex.. Solve: A linear decrease in velocity from t = 0 s to t = s implies a constant negative acceleration. On the other hand, a constant velocity between t = s and t = 4 s means zero acceleration.

13 .3. Visualize: Please refer to Figure Ex.3. Solve: (a) (b) The acceleration of the train at t = 3.0 s is + m/s.

14 .4. Visualize: Please refer to Figure Ex.4. Solve: (a) At t = s, the position of the particle is (b) From the graph itself at t = s, v = 4 m/s. (c) The acceleration is x s = m + area under velocity graph from t = 0 s to t = s = m + ( 4 m/s )( s ) = 6 m a x v v = = t v t 6 m/s 0 m/s = = m/s 3 s x fx ix

15 .5. Model: Represent the car as a particle. Solve: (a) First, we will convert units: The motion is constant acceleration, so 60 miles hour hour 60 m 3600 s mile = 6.8 m/s v v0 ( 6. 8 m/s 0 m/s) v = v0 + a t a = = =.68 m/s t 0 s (b) The fraction is a/g =.68/9.8 = So a is 7.3% of g. (c) The distance is calculated as follows: x = x0 + v0 t + a( t) = a( t) = 34 m = 440 feet

16 .6. Model: Represent the jet plane as a particle Visualize: Solve: (a) Since we don t know the time of acceleration, we will use v = v + a( x x ) v v0 ( 400 m/s) ( 300 m/s) a = = x ( 4000 m) 0 0 = m/s (b) The acceleration of the jet is approximately equal to g, the acceleration due to gravity.

17 .7. Model: We are using the particle model for the skater and the kinematics model of motion under constant acceleration. Solve: Since we don t know the time of acceleration we will use vf = vi + a( xf xi) vf vi ( 6.0 m/s) ( 8.0 m/s) a = = =.8 m/s ( x x ) ( 5.0 m) Assess: A deceleration of.8 m/s is reasonable. f i

18 .8. Model: Represent the spaceship as a particle. Solve: (a) The known information is: x 0 = 0 m, v 0 = 0 m/s, t 0 = 0 s, a = g = 9.8 m/s, and v = m/s. Constant acceleration kinematics gives v v0 7 v = v0 + a t t = t = = s a The problem asks for the answer in days, so we need a conversion: (b) The distance traveled is (c) The number of seconds in a year is: In one year light travels a distance hour day = (. s) = 354 days 3600 s 4 hour 7 t x x0 = v0 t + a( t) = at = m 4 hours 3600 s year = 365 days = s day hour light year = ( m/s) ( s) = m 5 5 The distance traveled by the spaceship is m/ m = of a light year. Assess: Note that x gives Where is it? rather than How far has it traveled? How far is represented by x x 0. They happen to be the same number in this problem, but that isn t always the case.

19 .9. Model: Represent the spherical drop of molten metal as a particle. Visualize: Solve: (a) The shot is in free fall, so we can use free fall kinematics with a = g. The height must be such that the shot takes 4 s to fall, so we choose t = 4 s. Then, y = y0 + v0( t t0) g( t t0) y0 = gt = ( 9.8 m/s )( 4 s) = 78.4 m (b) The impact velocity is v = v 0 g(t t 0 ) = gt = 39. m/s. Assess: Note the minus sign. The question asked for velocity, not speed, and the y-component of v r is negative because the vector points downward.

20 .0. Model: Assume the ball undergoes free fall acceleration and that the ball is a particle. Visualize: Solve: (a) We will use the kinematic equations v= v0 + at ( t0) and y= y0 + v0( t t0) + at ( t0) as follows: v (at t = s) = 9.6 m/s + ( 9.8 m/s )( s 0 s) = 9.8 m/s y (at t = s) = 0 m + (9.6 m/s)( s 0 s) + / ( 9.8 m/s )( s 0 s) = 4.7 m v (at t = s) = 9.6 m/s + ( 9.8 m/s )( s 0 s) = 0 m/s y (at t = s) = 0 m + (9.6 m/s)( s 0 s) + / ( 9.8 m/s )( s 0 s) = 9.6 m v (at t = 3 s) = 9.6 m/s + ( 9.8 m/s )(3 s 0 s) = 9.8 m/s y (at t = 3 s) = 0 m + (9.6 m/s)(3 s 0 s) + / ( 9.8 m/s )(3 s 0 s) = 4.7 m v (at t = 4 s) = 9.6 m/s + ( 9.8 m/s )(4 s 0 s) = 9.6 m/s y (at t = 4 s) = 0 m + (9.6 m/s)(4 s 0 s) + / ( 9.8 m/s )(4 s 0 s) = 0 m (b) Assess: (a) A downward acceleration of 9.8 m/s on a particle that has been given an initial upward velocity of +9.6 m/s will reduce its speed to 9.8 m/s after s and then to zero after s. The answers obtained in this solution are consistent with the above logic. (b) Velocity changes linearly with a negative uniform acceleration of 9.8 m/s. The position is symmetrical in time around the highest point which occurs at t = s.

21 .. Model: We represent the ball as a particle. Visualize: Solve: Once the ball leaves the student s hand, the ball undergoes free fall and its acceleration is equal to the acceleration due to gravity that always acts vertically downward towards the center of the earth. According to the constant-acceleration kinematic equations of motion Substituting the known values y = y0 + v0 t+ a t m = 0 m + ( 5 m/s) t + ( / )( 9.8 m/s ) t The solution of this quadratic equation gives t = 3. s. The other root of this equation yields a negative value for t, which is not valid for this problem. Assess: A time of 3. s is reasonable.

22 .. Model: We will use the particle model and the constant-acceleration kinematic equations. Visualize: Solve: (a) Substituting the known values into y = y + v t+ a t, we get m = 0 m + 0 (m/s) t + ( 9.8 m/s ) t One of the roots of this equation is negative and is not relevant physically. The other root is t = 4.53 s which is the answer to part (b). Using v = v0 + a t, we obtain v = 0( m/s) + ( 9.8 m/s )( 4.53 s ) = 4. 4 m/s (b) The time is 4.53 s. Assess: A time of 4.5 s is a reasonable value. The rock s velocity as it hits the bottom of the hole has a negative sign because of its downward direction. The magnitude of 4.4 m/s compared to 0 m/s, when the rock was tossed up, is consistent with the fact that the rock travels an additional distance of 0 m into the hole.

23 .3. Model: Represent the car as a particle. Visualize: Solve: 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 an accleration a = gsinθ = gsin 0 = 3.35 m/s. Constant acceleration kinematics gives v0 ( 30 m/s) v = v0 + a( x x0) 0 m/s = v0 + ax x = = = 34 m a ( 3.35 m/s ) Notice how the two negatives canceled to give a positive value for x. Assess: We must include the minus sign because the r a vector points down the slope, which is in the negative x-direction.

24 .4. Model: We will represent the skier as a particle. Visualize: Note that the skier s motion on the horizontal, frictionless snow is not of any interest to us. Also note that the acceleration parallel to the incline is equal to g sin0. Solve: Using the following constant-acceleration kinematic equations, vf x = vi x + ax ( xf xi) ( 5 m/s) = ( 3.0 m/s) + ( 9.8 m/s )sin 0 ( x 0 m ) x = m vf x = vi x + ax ( tf ti) ( 5 m/s) = ( 3.0 m/s) + ( 9.8 m/s )(sin 0 ) t t = s Assess: A time of 7.05 s to cover 63.5 m is a reasonable value.

25 3.5. Solve: (a) The position t = s is x s = [ ( ) + ] m = 5m. (b) The velocity is the derivative v = dx/ dt and the velocity at t = s is calculated as follows: (c) The acceleration is the derivative a v = ( 6t ) m/s v = [ 6( ) ] m/s = 3 m/s s = dv/ dt and the acceleration at t = s is calculated as follows: a = ( t) m/s a = [( )( )] m/s = 4 m/s s

26 .6. Solve: The formula for the particle s position along the x-axis is given by Using the expression for v x we get xf = xi + tf ti a 3 tf f = i + ti x x v dt dvx dt x d ( t m/s ) 4tm/s dt 3 3 [ ] x = = = 5 (a) The particle s position at t = s is x s = m + 3 m = 3 m. (b) The particle s speed at t = s is v s = m/s. (c) The particle s acceleration at t = s is a s = 4 m/s.

27 .7. Solve: The formula for the particle s velocity is given by v f = v i + area under the acceleration curve between t i and t f For t = 4 s, we get v 4 s = 8 m/s + m/s s = m/s Assess: The acceleration is positive but decreases as a function of time. The initial velocity of 8.0 m/s will therefore increase. A value of 6 m/s is reasonable.

28 .8. Solve: (a) Time (s) Position (m) (b) (c) s = s (at t = s) s (at t = 0 s) = 0 m ( 4 m) = 4 m. (d) s = s (at t = 4 s) s (at t = s) = 7 m 3 m = 4 m. (e) From t = 0 s to t = s, vs = s/ t = 4 m/s. (f) From t = s to t = 4 s, vs = s/ t = m/s. (g) The average acceleration is v m/s 4 m/s a = = = m/s t s s

29 .9. Solve: (a) (b) To be completed by student. dx (c) = vx = t 4 vx( at t = s ) = [ m/s ( s ) 4 m/s] = m/s dt (d) There is a turning point at t = s. (e) Using the equation in part (c), Since x = (t 4t + ) m, x = m. (f) v = x 4 m/s = ( t 4) m/s t = 4

30 .30. Visualize: Please refer to Figure P.30. Solve: (a) Using the equation, x f = x i + area under the velocity-versus-time graph between t i and t f we have, x (at t = s) = x (at t = 0 s) + area between t = 0 s and t = s =.0 m + (4 m/s)( s) = 6 m Reading from the velocity-versus-time graph, v x (at t = s) = 4 m/s. Also, a x = slope = v/ t = 0 m/s. (b) x (at t = 3.0 s) = x (at t = 0 s) + area between t = 0 s and t = 3 s =.0 m + 4 m/s s + m/s s + (/) m/s s = 3.0 m Reading from the graph, v x (t = 3 s) = m/s. The acceleration is vx( at t = 4 s) vx( at t = s) ax ( t = 3 s) = slope = = m/s s

31 .3. Visualize: Please refer to Figure P.3. Solve: The graph for particle A is a straight line from t = s to t = 8 s. The slope of this line is 0 m/s, which is the velocity at t = 7.0 s. The negative sign indicates motion toward lower values on the x-axis. The velocity of particle B at t = 7.0 s can be read directly from its graph. It is 0 m/s. The velocity of particle C can be obtained from the equation v f = v i + area under the acceleration curve between t i and t f This area can be calculated by adding up three sections. The area between t = 0 s and t = s is 40 m/s, the area between t = s and t = 5 s is 45 m/s, and the area between t = 5 s and t = 7 s is 0 m/s. We get (0 m/s) + (40 m/s) + (45 m/s) (0 m/s) = 75 m/s.

32 .3. Visualize: Please refer to Figure P.3. Solve: (a) We can calculate the position of the particle at every instant with the equation x f = x i + area under the velocity-versus-time graph between t i and t f The particle starts from the origin at t = 0 s, so x i = 0 m. Notice that the each square of the grid in Figure P.3 has area (5 m/s) ( s) = 0 m. We can find the area under the curve, and thus determine x, by counting squares. You can see that x = 35 m at t = 4 s because there are 3.5 squares under the curve. In addition, x = 35 m at t = 8 s because the 5 m represented by the half square between 4 and 6 s is cancelled by the 5 m represented by the half square between 6 and 8 s. Area beneath the axis are negative areas. The particle passes through x = 35 m at t = 4 s and again at t = 8 s. (b) The particle moves to the right for 0 s t 6 s, where the velocity is positive. It reaches a turning point at x = 40 m at t = 6 s. The motion is to the left for t > 6 s. This is shown in the motion diagram below.

33 .33. Visualize: Solve: We will determine the object s velocity using graphical methods first and then using calculus. Graphically, v(t) = v 0 + area under the acceleration curve from 0 to t. In this case, v 0 = 0 m/s. The area at each time t requested is a triangle. t = 0 s vt ( = 0 s) = v = 0 m/s t = s t = 4 s t = 6 s 0 vt ( = s) = ( s)( 5 m/s) = 5 m/s vt ( = 4 s) = ( 4 s)( 0 m/s) = 0 m/s vt ( = 6 s) = ( 6 s)( 0 m/s) = 30 m/s t = 8 s vt ( = 8 s) = vt ( = 6 s) = 30 m/s The last result arises because there is no additional area after t = 6 s. Let us now use calculus. The acceleration function a(t) consists of three pieces and can be written: 5. t 0 s t 4 s at () = 5t s t 6 s 0 6 s t 8 s These were determined by the slope and the y-intercept of each of the segments of the graph. The velocity function is found by integration as follows: For 0 t 4 s, This gives For 4 s t 6 s, This gives: For 6 s t 8 s, t t vt () = vt ( = 0 s) + atdt () = = 5. t t = 0 s t = s t = 4 s 0 vt ( = 0 s) = 0 m/s vt ( = s) = 5 m/s vt ( = 4 s) = 0 m/s t t vt () = vt ( = ) + atdt () = s 0 m/s + t. t t = t = 6 s vt ( = 6 s) = 30 m/s t vt () = vt ( = 6 s) + atdt () = 30 m/s + 0 m/s = 30 m/s 6 This gives: t = 8 s vt ( = 8 s) = 30 m/s Assess: The same velocities are found using calculus and graphs, but the graphical method is easier for simple graphs. t 0 t 4

34 .34. Visualize: Please refer to Figure P.34. Solve: (a) The velocity-versus-time graph is the derivative with respect to time of the distance-versus-time graph. The velocity is zero when the slope of the position-versus-time graph is zero, the velocity is most positive when the slope is most positive, and the velocity is most negative when the slope is most negative. The slope is zero at t = 0, s, s, 3 s,...; the slope is most positive at t = 0.5 s,.5 s,...; and the slope is most negative at t =.5 s, 3.5 s,... (b)

35 .35. Visualize: Please refer to Figure P.35. Solve: (a) We can determine the velocity as follows: v x = v x (at t = t i ) + area under the acceleration-versus-time graph from t = t i to t f v x (at t = s) = ( m/s )( s) = m/s v x (at t = 6 s) = m/s + 6 m/s = 4 m/s v x (at t = 4 s) = m/s + 0 m/s = m/s v x (at t = 8 s) = 4 m/s + m/s = 6 m/s v x (at t = 0 s) = 6 m/s 4 m/s = m/s (b) If v x (at t = 0 s) =.0 m/s, the entire velocity-versus-time graph will be displaced upward by.0 m/s.

36 .36. Model: Represent the ball as a particle. Visualize: The ball has a constant negative acceleration of 9.8 m/s until it makes contact with the floor. Velocity decreases linearly from zero to negative velocity before hitting the ground. Position decreases from the original height with a parabola to ground. Contact with the ground causes a large change in velocity toward up (positive velocity and large positive acceleration). The ball proceeds up with negative acceleration, slowing the ball and reversing the parabola of position. Solve:

37 .37. Solve: (a) The velocity-versus-time graph is given by the derivative with respect to time of the position function: dx v = x t t dt = ( 6 8 ) m/s For v x = 0 m/s, there are two solutions to the quadratic equation: t = 0 s and t = 3 s. (b) At the first of these solutions, x (at t = 0 s) = (0 s) 3 9(0 s) + = m The acceleration is the derivative of the velocity function: a x dvx = = ( t 8 ) m/s a ( at t = 0 s) = 8 m/s dt At the second solution, x (at t = 3 s) = (3 s) 3 9(3 s) + = 5 m a x (at t = 3 s) = (3 s) 8 = +8 m/s

38 .38. Model: Represent the object as a particle. Solve: (a) Known information: x 0 = 0 m, v 0 = 0 m/s, x = 40 m, v = m/s, t = 5 s. If the acceleration is uniform (constant a), then the motion must satisfy the three equations x = at a = 3.0 m/s v = at a =.0 m/s v = ax a =.5 m/s But each equation gives a different value of a. Thus the motion is not uniform acceleration. (b) We know two points on the velocity-versus-time graph, namely at t 0 = 0 and t = 5 s. What shape does the function have between these two points? If the acceleration was uniform, which it s not, then the graph would be a straight line. The area under the graph is the displacement x. From the figure you can see that x = 7.5 m for a straight-line graph. But we know that, in reality, x = 40 m. To get a larger x, the graph must bulge upward above the straight line. Thus the graph is curved, and it is concave downward.

39 .39. Solve: The position is the integral of the velocity. 3 x = x + v dt = x + kt dt = x + 3 kt = x + kt 0 t0 t t t We re given that x 0 = 9.0 m and that the particle is at x = 9.0 m at t = 3.0 s. Thus m = ( 90. m )+ 3 k( 30. s ) = 90. m k 90. s 3 3 Solving for k gives k = 0. m/s. x ( )+ ( )

40 .40. Solve: (a) The velocity is the integral of the acceleration. t t t x 0x x t ( ) = ( ) = v = v + a dt = 0 m/s+ 0 t dt 0t t 0t t The velocity is zero when vx = 0 m/s = ( 0t t )= ( 0 t ) t t = 0 s or t = 0 s The first solution is the initial conditions. Thus the particle s velocity is again 0 m/s at t = 0 s. (b) Position is the integral of the velocity. At t = 0 s, and using x 0 = 0 m at t 0 = 0 s, the position is t 3 x = x0 + vxdt = 0 m + 0t t dt 5t 6 t 667 m t ( ) = = 0 0

41 .4. Model: Represent the ball as a particle. Visualize: Please refer to Figure.4. Solve: In the first and third segments the acceleration a s is zero. In the second segment the acceleration is negative and constant. This means the velocity v s will be constant in the first two segments and will decrease linearly in the third segment. Because the velocity is constant in the first and third segments, the position s will increase linearly. In the second segment, the position will increase parabolically rather than linearly because the velocity decreases linearly with time.

42 .4. Model: Represent the ball as a particle. Visualize: Please refer to Figure P.4. The ball rolls down the first short track, then up the second short track, and then down the long track. s is the distance along the track measured from the left end (where s = 0). Label t = 0 at the beginning, that is, when the ball starts to roll down the first short track. Solve: Because the incline angle is the same, the magnitude of the acceleration is the same on all of the tracks. Assess: Note that the derivative of the s versus t graph yields the v s versus t graph. And the derivative of the v s versus t graph gives rise to the a s versus t graph.

43 .43. Model: Represent the ball as a particle. Visualize: Please refer to Figure P.43. The ball moves to the right along the first track until it strikes the wall which causes it to move to the left on a second track. The ball then descends on a third track until it reaches the fourth track which is horizontal. Solve: Assess: Note that the time derivative of the position graph yields the velocity graph, and the derivative of the velocity graph gives the acceleration graph.

44 .44. Visualize: Please refer to Figure P.44. Solve:

45 .45. Visualize: Please refer to Figure P.45. Solve:

46 .46. Visualize: Please refer to Figure P.46. There are four frictionless tracks. Solve: For the first track, a s is negative and constant and v s is decreasing linearly. This is consistent with a ball rolling up a straight track but not so far that v s goes to zero. For the second track, a s is zero and v s is constant but greater than zero. This is consistent with a ball rolling on a horizontal track. For the third track, a s is positive and constant and v s is increasing linearly. This is consistent with a ball rolling down a straight track. For the fourth track, a s is zero and v s is constant. This is again consistent with a ball rolling on a horizontal track. The a s on the first track has the same absolute value as the a s on the third track. This means the slope of the first track up is the same as the slope of the third track down.

47 .47. Model: The plane is a particle and the constant-acceleration kinematic equations hold. Solve: (a) To convert 80 m/s to mph, we calculate 80 m/s mi/609 m 3600 s/hr = 79 mph. (b) Using a = s v/ t, we have, 3 m/s 0 m/s 69 m/s 46 m/s as( t = 0 to t = 0 s) = =.3 m/s as( t = 0 s to t = 30 s) = =.3 m/s 0 s 0 s 30 s 0 s For all time intervals a is.3 m/s. (c) Using kinematics as follows: v = v + a( t t ) 80 m/s = 0 m/s + (.3 m/s )( t 0 s) t = 35 s fs is f i (d) Using the above values, we calculate the takeoff distance as follows: sf = si + vis( tf ti) + as( tf ti) = 0 m + ( 0 m/s)( 35 s) + (.3 m/s )( 35 s) = 40 m For safety, the runway should be 3 40 m = 430 m or.6 mi. This is longer than the.4 mi long runway, so the takeoff is not safe. f f

48 .48. Model: The automobile is a particle. Solve: (a) The acceleration is not constant because the velocity-versus-time graph is not a straight line. (b) Acceleration is the slope of the velocity graph. You can use a straightedge to estimate the slope of the graph at t = s and at t = 8 s. Alternatively, we can estimate the slope using the two data points on either side of s and 8 s. Either way, you need the conversion factor mph = m/s from Table.3. vx( at 4 s) vx( at 0 s) mph m/s ax ( at s) =. 5 = 5. m/s 4 s 0 s s mph ( ) ( ) vx at 0 s vx at 6 s ax ( at 8 s) 0 s 6 s (c) The displacement, or distance traveled, is 0 s mph m/s = 45. = 0. m/s s mph x = x( at 0 s) x( at 0 s)= v dx = area under the velocity curve from 0 s to 0 s 0 s x We can approximate the area under the curve as the area of five rectangular steps, each with width t = s and height equal to the average of the velocities at the beginning and end of each step. 0 s to s vavg = 4mph = 66. m/s area = 5. m s to 4 s vavg = 37 mph = 6. 5 m/s area = m 4 s to 6 s vavg = 53 mph = 3. 7 m/s area = m 6 s to 8 s vavg = 65 mph = 9. m/s area = 58. m 8 s to 0 s v = 74 mph = 33. m/s area = 66. m The total area under the curve is 7 m, so the distance traveled in 0 s is 7 m. avg

49 .49. Model: The car is a particle, and constant-acceleration kinematic equations hold. Visualize: Solve: This is a two-part problem. During the reaction time, we can use x = x0 + v0( t t0) + a0( t t0) = 0 m + (0 m/s)(0.50 s 0 s) + 0 m = 0 m During deceleration, v = v + a ( x x ) 0 = (0 m/s) + ( 6.0 m/s )(x 0 m) x = 43.3 m She has 50 m to stop, so she can stop in time.

50 .50. Model: The car is a particle and constant-acceleration kinematic equations hold. Visualize: Solve: (a) This is a two-part problem. During the reaction time, x = x 0 + v 0 (t t 0 ) + / a 0 (t t 0 ) = 0 m + (0 m/s)(0.50 s 0 s) + 0 m = 0 m After reacting, x x = 0 m 0 m = 00 m, that is, you are 00 m away from the intersection. (b) To stop successfully, v = v + a ( x x ) ( 0 m/s) = ( 0 m/s) + a ( 00 m) a = m/s (c) The time it takes to stop can be obtained as follows: v = v + a ( t t ) 0 m/s = 0 m/s + ( m/s )( t 050. s) t = 0.5 s

51 .5. Model: The car is represented as a particle. Visualize: Solve: (a) This is a two-part problem. First, we need to use the information given to determine the acceleration during braking. Second, we need to use that acceleration to find the stopping distance for a different initial velocity. First, the car coasts at constant speed before braking: x = x + v ( t t ) = v t = ( 30 m/s)( 0.5 s) = 5 m Then, the car brakes to a halt. Because we don t know the time interval, use v = 0 = v + a ( x x ) v ( 30 m/s) a = = = 0 m/s ( x x ) ( 60 m 5 m) We use v = v 0 = 30 m/s. Note the minus sign, because a r points to the left. We can repeat these steps now with v 0 = 40 m/s. The coasting distance before braking is x = v t = ( 40 m/s)( 0.5 s) = 0 m The braking distance is 0 v = 0 = v + a ( x x ) v ( 40 m/s) x = x = 0 m = 00 m a ( 0 m/s ) (b) The car coasts at a constant speed for 0.5 s, traveling 0 m. The graph will be a straight line with a slope of 40 m/s. For t 0.5 the graph will be a parabola until the car stops at t. We can find t from v v = 0 = v + a( t t) t = t = 4.5 s a The parabola will reach zero slope (v = 0 m/s) at t = 4.5 s. This is enough information to draw the graph shown in the figure.

52 .5. Model: We will use the particle model and the constant-acceleration kinematic equations. Visualize: Solve: (a) To find x, we first need to determine x. Using x = x 0 + v 0 (t t 0 ), we get x = 0 m + (0 m/s)(0.50 s 0 s) = 0 m. Now, v = v + a ( x x ) 0 m /s = ( 0 m/s ) + ( 0 m/s )( x 0 m) x = 30 m The distance between you and the deer is (x 3 x ) or (35 m 30 m) = 5 m. (b) Let us find v 0 max such that v = 0 m/s at x = x 3 = 35 m. Using the following equation, v v = a ( x x ) 0 m /s v = ( 0 m/s )( 35 m x ) 0 max 0 max Also, x = x 0 + v 0 max (t t 0 ) = v 0 max (0.50 s 0 s) = (0.50 s)v 0 max. Substituting this expression for x in the above equation yields v = ( 0 m/s )[ 35 m ( s ) v ] v + ( 0 m/s ) v 700 m /s = 0 0 max 0 max 0 max 0 max The solution of this quadratic equation yields v 0 max =.9 m/s. (The other root is negative and unphysical for the present situation.) Assess: An increase of speed from 0 m/s to.9 m/s is very reasonable for the car to cover an additional distance of 5 m with a reaction time of 0.50 s and a deceleration of 0 m/s.

53 .53. Model: The rocket is represented as a particle. Visualize: Solve: (a) There are three parts to the motion. Both the second and third parts of the motion are free fall, with a = g. The maximum altitude is y. In the acceleration phase: y = y0 + v0( t t0) + a( t t0) = at = ( 30 m/s )( 30 s) = 3,500 m v = v + a( t t ) = at = ( 30 m/s )( 30 s) = 900 m/s 0 0 In the coasting phase, v ( 900 m/s) v = 0 = v g( y y) y = y+ = 3,500 m + = 54,800 m = 54.8 km g ( 9.8 m/s ) The maximum altitude is 54.8 km ( 33 miles). (b) The rocket is in the air until time t 3. We already know t = 30 s. We can find t as follows: v v = 0 m/s = v g( t t) t = t+ = s g Then t 3 is found by considering the time needed to fall 54,800 m: y y3 = 0 m = y + v( t3 t) g( t3 t) = y g( t3 t) t3 = t + = 8 s g (c) The velocity increases linearly, with a slope of 30 (m/s)/s, for 30 s to a maximum speed of 900 m/s. It then begins to decrease linearly with a slope of 9.8 (m/s)/s. The velocity passes through zero (the turning point at y ) at t = s. The impact velocity at t 3 = 8 s is calculated to be v 3 = v g(t 3 t ) = 040 m/s. Assess: In reality, friction due to air resistance would prevent the rocket from reaching such high speeds as it falls, and the acceleration upward would not be constant because the mass changes as the fuel is burned, but that is a more complicated problem.

54 .54. Model: We will model the rocket as a particle. Air resistance will be neglected. Visualize: Solve: (a) Using the constant-acceleration kinematic equations, v = v0 + a0( t t0) = 0 m/s + a0( 6 s 0 s ) = a0( 6 s) y = y0 + v0( t t0) + a0( t t0) = a0( 6 s 0 s ) = a0( 8 s ) y = y + v( t t) + a( t t) 500 m = 8 s a0 + 6 s a0( 0 s 6 s ) + ( 9. 8 m/s )( 0 s 6 s ) a 0 = 7 m/s (b) The rocket s speed as it passes through a cloud 500 m above the ground can be determined using the kinematic equation: v = v + a (t t ) = (6 s) a 0 + ( 9.8 m/s )(4 s) = 39 m/s Assess: 400 m/s 900 mph, which is reasonable for a rocket that has been accelerating for 0 s at a rate of approximately 54 mps/s or 4 m/s.

55 .55. Model: We will model the lead ball as a particle and use the constant-acceleration kinematic equations. Visualize: Note that the particle undergoes free fall until it hits the water surface. Solve: The kinematics equation y = y + v ( t t ) + a ( t t ) becomes Now, once again, m = 0 m + 0 m + ( 98. m/s )( t 0) t =.0 s y = y+ v( t t) + a( t t) y y = v ( 3.0 s.0 s ) + 0 m/s =.99 v v is easy to determine since the time t has been found. Using v = v + a ( t t ), we get With this value for v, we go back to: v = 0 m/s ( 9.8 m/s )(.0 s 0 s) = m/s y y =. 99v = (. 99)( m/s) = 9.7 m y y is the displacement of the lead ball in the lake and thus corresponds to the depth of the lake. The negative sign shows the direction of the displacement vector. Assess: A depth of about 60 ft for a lake is not unusual.

56 .56. Model: The elevator is a particle moving under constant-acceleration kinematic equations. Visualize: Solve: (a) To calculate the distance to accelerate up: ( v ) = v + a ( y y ) ( 5 m/s ) = ( 0 m/s) + ( m/s )( y 0 m) y =.5 m (b) To calculate the time to accelerate up: v = v + a ( t t ) 5 m/s = 0 m/s + ( m/s )( t 0 s) t = 5 s To calculate the distance to decelerate at the top: v = v + a ( y y ) ( 0 m/s) = ( 5 m/s ) + ( m/s )( y y ) y y =.5 m 3 3 To calculate the time to decelerate at the top: The distance moved up at 5 m/s is 3 3 v = v + a ( t t ) 0 m/s = 5 m/s+ ( m/s )( t t ) t t = 5 s y y = ( y y ) ( y y ) ( y y ) = 00 m.5 m.5 m = 75 m The time to move up 75 m is given by y y = v( t t) + a( t t) 75 m = ( 5 m/s) ( t t) ( t t) = 35 s To total time to move to the top is ( t t ) + ( t t ) + ( t t ) = 5 s + 35 s + 5 s = 45 s 0 3 Assess: To cover a distance of 00 m at 5 m/s (ignoring acceleration and deceleration times) will require a time of 40 s. This is comparable to the time of 45 s for the entire trip as obtained above.

57 .57. Model: The car is a particle moving under constant-acceleration kinematic equations. Visualize: Solve: This is a three part problem. First the car accelerates, then it moves with a constant speed, and then it decelerates. First, the car accelerates: v = v0 + a0( t t0) = 0 m/s + ( 4.0 m/s )( 6 s 0 s) = 4 m/s x = x0 + v0( t t0) + a0( t t0) = 0 m + ( 4.0 m/s )( 6 s 0 s) = 7 m Second, the car moves at v : x x = v( t t) + a( t t) = ( 4 m/s)( 8 s 6 s ) + 0 m = 48 m Third, the car decelerates: v3 = v + a( t3 t) 0 m/s = 4 m/s + ( 3.0 m/s )( t3 t) ( t3 t) = 8 s x3 = x + v( t3 t) + a( t3 t) x3 x = ( 4 m/s)( 8 s) + ( 3.0 m/s )( 8 s) = 96 m Thus, the total distance between stop signs is: x x = ( x x ) + ( x x ) + ( x x ) = 96 m + m + 7 m = 6 m Assess: A distance of approximately 600 ft in a time of around 0 s with an acceleration/deceleration of the order of 7 mph/s is reasonable.

58 .58. Model: The car is a particle moving under standard equations of motion. Visualize: Solve: Using the kinematic equation for position: x = x + v( t t) + a( t t) x + 30 m = x + v( 5.0 s 4.0 s) + ( m/s )( 5.0 s 4.0 s) 30 m = v(.0 s) + ( m/s )(.0 s) v = 9 m/s And 4.0 seconds before: v = v + a ( t t ) 9 m/s = v + ( m/s )( 4.0 s 0 s) v = m/s Assess: m/s 47 mph and is a reasonable value.

59 .59. Model: Santa is a particle moving under constant-acceleration kinematic equations. Visualize: Note that our x-axis is positioned along the incline. Solve: Using the following kinematic equation, v = v0 + a ( x x ) = ( 0 m/s ) + ( 4.9 m/s )( 0 m 0 m ) v = 9.9 m/s Assess: Santa s speed of 0 mph as he reaches the edge is reasonable.

60 .60. Model: The cars are represented as particles. Visualize: Solve: (a) Ann and Carol start from different locations at different times and drive at different speeds. But at time t they have the same position. It is important in a problem such as this to express information in terms of positions (that is, coordinates) rather than distances. Each drives at a constant velocity, so using constant velocity kinematics gives xa = xa0 + va ( t ta0 ) = va ( t ta0 ) xc = xc0 + vc( t tc0 ) = xc0 + vct The critical piece of information is that Ann and Carol have the same position at t, so x A = x C. Equating these two expressions, we can solve for the time t when Ann passes Carol: v ( t t ) = x + v t A A0 C0 C ( va vc ) t = xc0 + vata0 xc0 + vata0.4 mi + ( 50 mi/hr) ( 0.5 hr) t = = =.957 hours va vc 50 mi/hr 36 mi/hr (b) Their position is x = x A = x C = x C0 v C t = 7.9 miles. (c) Note that Ann s graph doesn t start until t = 0.5 hours, but her graph has a steeper slope so it intersects Carol s graph at t =.957 hours.

61 .6. Model: We will use the particle model for the ball. Visualize: 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 axes point in different directions, we can avoid possible confusion by calling the downhill axis the z-axis and the downhill velocities u. The uphill axis as usual will be denoted by x and the uphill velocities as v. Note that the height information, h = m, has to be transformed into information about positions along the two axes. Solve: (a) The uphill roll has a = g sin α = 6.93 m/s. The speed at the top is found from 0 v = v + a ( x x ) v = v + a x = ( 5 m/s) + ( 6.93 m/s )(.44 m) =.3 m/s 0 0 (b) The downward roll starts with velocity 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 = 5.00 m/s (c) The final speed is equal to the initial speed, so the percentage change is zero! This result may seem surprising, but can be more easily understood after we introduce the concept of energy.

62 .6. Model: We will model the toy train as a particle. Visualize: Solve: Using kinematics, x = x0 + v0( t t0) + a0( t t0) = m + (.0 m/s)(.0 s 0 s) + 0 m = 6.0 m The acceleration can now be obtained as follows: v = v + a ( x x ) 0 m /s = (.0 m/s) + a ( 8.0 m 6.0 m) a =.0 m/s Assess: A deceleration of m/s in bringing the toy car to a halt over a distance of.0 m is reasonable.

63 .63. Model: We will use the particle model and the kinematic equations at constant-acceleration. Visualize: Solve: To find x, let us use the kinematic equation v = v + a ( x x ) = ( 0 m/s) = ( 50 m/s) + ( 0 m/s )( x x ) x = x + 5 m Since the nail strip is at a distance of 50 m from the origin, we need to determine x : x = x + v ( t t ) = m + ( 50 m/s)( 0.60 s 0.0 s) = 30 m Therefore, we can see that x = (30 + 5) m = 55 m. That is, he can t stop within a distance of 50 m. He is in jail. Assess: Bob is driving at approximately 00 mph and the stopping distance is of the correct order of magnitude.

64 .64. Model: We will use the particle model with constant-acceleration kinematic equations. Visualize: Solve: The acceleration, being the same along the incline, can be found as v = v + a( x x ) ( 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 + a( t 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 x as follows: x = x0 + v0( t t0) + a( t t0) = 0 m + ( 5.0 m/s)( 3.3 s) + (.5 m/s )( 3.3 s) = 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 x 0 and the puck stops 0. m or 0 cm before the ramp ends, you are not a winner.

65 .65. Model: We will use the particle model for the skier s motion ignoring air resistance. Visualize: Solve: (a) As discussed in the text, acceleration along a frictionless incline is a = gsin 5, where g is the acceleration due to gravity. The acceleration of the skier on snow therefore is a = ( 090. ) gsin 5. Also since h/ x = sin 5, x = 00 m/sin5. The final velocity can now be determined using kinematics v = v0 + a x x0 = m ( ) ( m/s) (. )( 9.8 m/s )sin 0 5 m sin v = = km ( 3600 s m/s m/s) 000 m hr = 4 km/hr (b) The speed lost to air resistance is ( 4 80)/ 4 00% = 6%. Assess: A record of 80 km/hr on such a slope in the presence of air resistance makes the obtained speed of 4 km/hr (without air resistance) physical and reasonable.

66 .66. Model: We will use the particle model for the motion of the rocks which move according to constantacceleration kinematic equations. Visualize: Solve: (a) For Heather, yh = yh0 + vh0( th th0 ) + a0( th th0 ) 0 m = ( 50 m) + ( 0 m/s)( th 0 s) + ( 9.8 m/s )( th 0 s) 4. 9 m/s t + 0 m/s t 50 m = 0 H The two mathematical solutions of this equation are 5.83 s and +.75 s. The first value is not physically acceptable since it represents a rock hitting the water before it was thrown, therefore, t H =.75 s. For Jerry, yj = yj0 + vj0( tj tj0) + a0( tj tj0) 0 m = ( 50 m) + ( + 0 m/s )( tj 0 s) + ( 9.8 m/s )( tj 0 s) Solving this quadratic equation will yield t J =.75 s and s. Again only the positive root is physically meaningful. The elapsed time between the two splashes is t J t H = 5.83 s.75 s = 4.08 s. (b) Knowing the times, it is easy to find the impact velocities: vh = vh0 + a0( th th0 ) = ( 0 m/s) + ( 9.8 m/s)(.75 s 0 s) = 37. m/s vj = vj0 + a0( tj tj0 ) = ( + 0 m/s) + ( 9.8 m/s )( 5.83 s 0 s) = 37. m/s Assess: The two rocks hit water with equal speeds. This is because Jerry s rock has the same downward speed as Heather s rock when it reaches Heather s starting position during its downward motion. H

67 .67. Model: The ball is a particle that exhibits freely falling motion according to the constant-acceleration kinematic equations. Visualize: Solve: Using the known values, we have v = v + a ( y y ) ( 0 m/s) = v + ( 9.8 m/s )( 5.0 m 0 m) v = 4 m/s

68 .68. Model: The car is a particle that moves under the standard equations of motion. Visualize: Solve: The reaction time is.0 s, and the motion during this time is x = x + v ( t t ) = m + ( 0 m/s)(.0 s) = 0 m During slowing down, x = x+ v( t t) + a( t t) = 00 m = 0 m + ( 0 m/s) ( 5 s.0 s) + a( 5 s.0 s) a =.00 m/s The final speed v can now be obtained as v = v + a ( t t ) = ( 0 m/s) + (.00 m/s )( 5 s s ) =. m/s 57

69 .69. Model: Both cars are particles that move according to the constant-acceleration kinematic equations. Visualize: Solve: (a) David s and Tina s motions are given by the following equations: xd = xd0 + vd0( td td0) + ad( td td0) = vd0td x = x + v ( t t ) + a ( t t ) = 0 m + 0 m + a t T T0 T0 T T0 T T T0 T T When Tina passes David the distances are equal and t D = t T, so we get Using Tina s position equation, vd0 ( 30 m/s) xd = xt vd0td = attt vd0 = attt tt = = = 30 s a.0 m/s xt = attt = (.0 m/s )( 30 s) = 900 m (b) Tina s speed v T can be obtained from v = v + a ( t t ) = ( 0 m/s) + (.0 m/s )( 30 s 0 s) = 60 m/s T T0 T T T0 Assess: This is a high speed for Tina (~34 mph) and so is David s velocity (~67 mph). Thus the large distance for Tina to catch up with David (~0.6 miles) is reasonable. T

70 .70. Model: We are assuming both cars are particles. Visualize: Solve: The Porsche s time to finish the race is determined from the position equation xp = xp0 + vp0( tp tp0 ) + ap ( tp tp0 ) 400 m = 0 m + 0 m + ( 3.5 m/s )( tp 0 s) tp = 5. s The Honda s time to finish the race is obtained from Honda s position equation as xh = xh0 + vh0( th th0 ) + ah0( th th0 ) 400 m = 50 m + 0 m + ( 3.0 m/s )( th 0 s) th = 5.3 s So, the Porsche wins.

71 .7. Model: We will represent the dog and the cat in the particle model. Visualize: Solve: We will first calculate the time t C the cat takes to reach the window. The dog has exactly the same time to reach the cat (or the window). Let us therefore first calculate t C as follows: xc = xc0 + vc0( tc tc0 ) + ac( tc tc0 ) 3.0 m =.5 m + 0 m + ( 0.85 m/s ) tc tc =.879 s In the time t D =.879 s, the dog s position can be found as follows: xd = xd0 + vd0( td td0 ) + ad( td td0 ) = 0 m + (.50 m/s)(.879 s) + ( 0.0 m/s )(.879 s) =.6 m That is, the dog is shy of reaching the cat by 0.4 m. The cat is safe.

72 .7. Model: We use the particle model for the large train and the constant-acceleration equations of motion. Visualize: Solve: Your position after time t Y is xy = xy0 + vy0( ty ty0 ) + ay( ty ty0 ) = 0 m + ( 8.0 m/s )( t 0 s) + 0 m = 8.0 t The position of the train, on the other hand, after time t T is xt = xt0 + vt0( tt tt0 ) + at ( tt tt0 ) = 30 m + 0 m + (.0 m/s )( tt) = t The two positions x Y and x T will be equal at time t Y (= t T ) if you are able to jump on the back step of the train. That is, t = 8. 0t t ( 6 s ) t + 60 s = 0 t = 6 s and 0 s Y Y Y The first time, 6 s, is when you will overtake the train. If you continue to run alongside, the accelerating train will then pass you at 0 s. Let us now see if the first time t Y = 6.0 s corresponds to a distance before the barrier. From the position equation for you, x Y = ( 8.0 m/s)( 6.0 s) = 48.0 m. The position equation for the train will yield the same number. Since the barrier is at a distance of 50 m from your initial position, you can just catch the train before crashing into the barrier. Y Y Y Y T

73 .73. Model: The rocket and the bolt will be represented as particles to investigate their motion. Visualize: The initial velocity of the bolt as it falls off the side of the rocket is the same as that of the rocket, that is, v B0 = v R and it is positive since the rocket is moving upward. The bolt continues to move upward with a deceleration equal to g = 9.8 m/s before it comes to rest and begins its downward journey. Solve: To find a R we look first at the motion of the rocket: yr = yr0 + vr0( tr tr0 ) + ar ( tr tr0 ) = 0 m + 0 m/s + ar( 4.0 s 0 s) = 8a To find a R we must determine the magnitude of y R or y B0. Let us now look at the bolt s motion: yb = yb0 + vb0( tb tb0 ) + ab( tb tb0 ) 0 = yr + vr( 6.0 s 0 s) + ( 9.8 m/s )( 6.0 s 0 s) y = 76.4 m (6.0 s) v R Since vr = vr0 + ar ( tr tr0 ) = 0 m/s+ 4 ar = 4 ar the above equation for y R yields y R = (4a R ). We know from the first part of the solution that y R = 8a R. Therefore, 8a R = a R and hence a R = 5.5 m/s. R R

74 .74. Solve: A comparison of the given equation with the constant-acceleration kinematics equation x = x0 + v0( t t0) + ax( t t0) yields the following information: x 0 = 0 m, x = 64 m, t 0 = 0, t = 4 s, and v 0 = 3 m/s. (a) After landing on the deck of a ship at sea with a velocity of 3 m/s, a fighter plane is observed to come to a complete stop in 4.0 seconds over a distance of 64 m. Find the plane s deceleration. (b) (c) 64 m = 0 m + ( 3 m/s) ( 4 s 0 s) + ax( 4 s 0 s ) 64 m = 8 m + ( 8 s ) ax ax = 8 m/s

75 .75. Solve: (a) A comparison of this equation with the constant-acceleration kinematic equation ( v ) = v + ( a )( y y ) y 0y yields the following information: y 0 = 0 m, y = 0 m, a y = 9.8 m/s, and v y = 0 m/s. It is clearly a problem of free fall. On a romantic Valentine Day, John decided to surprise his girlfriend, Judy, in a special way. As he reached her apartment building, he found her sitting in the balcony of her second floor apartment 0.0 m above the first floor. John quietly armed his spring-loaded gun with a rose, and launched it straight up to catch her attention. Judy noticed that the flower flew past her at a speed of 0 m/s. Judy is refusing to kiss John until he tells her the initial speed of the rose as it was released by the spring-loaded gun. Can you help John on this Valentine Day? (b) y 0 (c) ( 0 m/s) = v0y ( 9.8 m/s )( 0 m 0 m) v0y = 7. m/s Assess: The initial velocity of 7. m/s, compared to a velocity of 0 m/s at a height of 0 m, is very reasonable.

76 .76. Solve: A comparison with the constant-acceleration kinematics equation ( v ) = ( v ) + a ( x x ) x 0x x yields the following quantities: x 0 = 0 m, v 0x = 5 m/s, v x = 0 m/s, and a x = (9.8 m/s )sin0. (a) A wagon at the bottom of a frictionless 0 incline is moving up at 5 m/s. How far up the incline does it move before reversing direction and then rolling back down? (b) 0 (c) ( 0 m/s) = ( 5 m/s) ( 9.8 m/s )sin 0 ( x 0 m) 5( m/s ) = ( 9. 8 m/s )( 0. 74) x x = 7.3 m

77 .77. Solve: (a) From the first equation, the particle starts from rest and accelerates for 5 s. The second equation gives a position consistent with the first equation. The third equation gives a subsequent position following the second equation with zero acceleration. A rocket sled accelerates from rest at 0 m/s for 5 sec and then coasts at constant speed for an additional 5 sec. Draw a graph showing the velocity of the sled as a function of time up to t = 0 s. Also, how far does the sled move in 0 s? (b) (c) x = ( 0 m/s )( 5 s) = 50 m v = 0 m/s ( 5 s) = 00 m/s x = 50 m + ( 00 m/s) ( 5 s) = 750 m

78 .78. Model: Jill and the grocery cart will be treated as particles that move according to the constantacceleration kinematic equations. Visualize: Solve: The final position of Jill when the cart is caught is given by xj = xj0 + vj0( tj tj0 ) + aj0( tj tj0 ) = 0 m + 0 m + aj0( tj 0 s) = (.0 m/s ) t The cart s position when it is caught is J xc = xc0 + vc0( tc tc0 ) + ac0( tc tc0 ) = 50 m + 0 m + ( 0.5 m/s )( tc 0 s) = 50 m + ( 0.5 m/s ) t Since x J = x C and t J = t C, we get C (. 0) tj = 50 s tC 0. 75tC = 50 s tc = 8.0 s x = 50 m + ( 0.5 m/s ) t = 50 m + ( 0. 5 m/s )( 8. s) = 67. m C So, the cart has moved 7. m. C

79 .79. Model: The watermelon and Superman will be treated as particles that move according to constantacceleration kinematic equations. Visualize: Solve: The watermelon s and Superman s position as they meet each other are Because t S = t W, Since y W = y S, yw = yw0 + vw0( tw tw0) + aw0( tw tw0) ys = ys0 + vs0( ts ts0) + as0( ts ts0) yw = 30 m + 0 m + ( 9.8 m/s )( tw 0 s) y = 30 m + ( 35 m/s) ( t 0 s) + 0 m S y = 30 m (4.9 m/s ) t y = 30 m (35 m/s) t W W 30 m ( 4. 9 m/s ) t = 30 m ( 35 m/s) t t = 0 s and 7. s W Indeed, t W = 0 s corresponds to the situation when Superman arrives just as the watermelon is dropped off the Empire State Building. The other value, t W = 7. s, is the time when the watermelon will catch up with superman. The speed of the watermelon as it passes Superman is v = v + a ( t t ) = 0 m/s + ( 9.8 m/s )( 7. s 0 s) = 70 m/s W W0 W0 W W0 Note that the negative sign implies a downward velocity. Assess: A speed of 40 mph for the watermelon is understandable in view of the significant distance (50 m) involved in the free fall. S S W W W

80 .80. Model: The rocket car is a particle that moves according to the constant-acceleration equations of motion. Visualize: Solve: This is a two part problem. For the first part, x = x0 + v0( t t0) + a0( t t0) = 0 m + 0 m + a0( 9.0 s 0 s) = ( 8 s ) a0 v = v0 + a0( t t0) = 0 m/s + a0( 9.0 s 0 s) = ( 9.0 s) a0 During the second part of the problem, x = x + v( t t) + a( t t) 990 m = ( 8 s ) a0 + ( 9.0 s) a0( s 9.0 s) + ( 5.0 m/s )( s 9.0 s) a = 5 m/s This leads to: v 0 = ( 9.0 s) a = ( 9.0 s)( 5 m/s ) = 35 m/s 0 Using this value of v, we can now calculate v as follows: v = v + a (t t ) = (35 m/s) + ( 5.0 m/s )( s 9.0 s) = 0 m/s That is, the car s speed as it passes the judges is 0 m/s. Assess: This is a very fast motion (~50 mph), but the acceleration is large and the long burn time of 9 s yields a high velocity.

81 .8. Model: Use the particle model. Visualize: Solve: (a) Substituting into the constant-acceleration kinematic equation 0 x = x+ v( t t) + a( t t) 00 m = x+ v t m 00 x 0 t = + v 3 Let us now find v and x as follows: 0 v = v0 + a0( t t0) = 0 m/s + ( 3.6 m/s ) s 0 s 3 = m/s 0 x = x0 + v0( t t0) + a0( t t0) = 0 m + 0 m + ( 3.6 m/s ) s 0 s = 0 m 3 The expression for t can now be solved as 00 m 0 m 0 s t = + = 0 s m/s 3 (b) The top speed = m/s which means v = m/s. To find the acceleration so that the sprinter can run the 00- meter dash in 9.9 s, we use m/s v = v0 + a0( t t0) m/s = 0 m/s + a0t t = a0 x = x0 + v0( t t0) + a0( t t0) = 0 m + 0 m + at 0 = at 0 Since x = x + v ( t t ) + a ( t t ), we get Substituting the above equation for t in this equation, 00 m = at 0 + ( m/s ) ( 9.9 s t) + 0 m m/s m/s 00 m = m/s 9.9 s 3.83 m/s a + ( ) a a a = (c) We see from parts (a) and (b) that the acceleration has to be increased from 3.6 m/s to 3.83 m/s for the sprint time to be reduced from 0 s to 9.9 s, that is, by %. This decrease of time by % corresponds to an increase of acceleration by % = 6. 4% 36. 0