KINEMATICS REVIEW VECTOR ALGEBRA - SUMMARY

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1 1 KINEMATICS REVIEW VECTOR ALGEBRA - SUMMARY Magnitude A numerical value with appropriate units. Scalar is a quantity that is completely specified by magnitude. Vector requires both, magnitude and direction for a complete description. The main difference between scalars and vectors is difference between scalar and vector algebra geometrical description of vector quantity: the direction of a vector is the counterclockwise angle of rotation which that vector makes with positive direction of x - axis. Two vectors are equal if they have the same magnitude and directions. Vectors that have the same magnitude and the same direction are the same. Multiplying a vector by a scalar: Multiplying vector by increases its magnitude by a factor. Opposite vectors: AA aaaaaa AA Multiplying by ½ changes the magnitude ½ times and reverses the direction Components of a Vector (x or horizontal component and y or vertical component) Any vector can be resolved into two components. If you know the magnitude A and direction θ of a vector A = (A, θ) you can find x and y components of that vector: A = Acosθ if (A,θ) A = Asinθ known x y If you know x and y components of a vector A you can find the magnitude A and direction θ of that vector: A= A +A x y if (A x,a y) Ay θ=arc tan known A x if the vector is in the first quadrant; if not you find it from the picture. v = Find v x and v y v x = (34 m/s )cos 48 ) = 3 m/s; v y = (34 m/s) sin 48 = 5 m/s F x = 4 N and F y = 3 N. Find magnitude (always positive!) and direction. F = 4 +3 = 5N ; θ = arc tan (¾) = 37 0 Vector addition graphically comparison between head-to-tail and parallelogram method Two methods for vector addition are equivalent. examples: Find FF 11 + FF = FF using both methods. The sum is the vector sum of the two individual vectors, known as the "resultant" or net vector " examples: resultant velocity, resultant force or net force (the one that can replace all forces that are applied at the same time)

2 SUBTRACTION is adding opposite vector. C = A - B = A + (-B) ANALYTICALLY/NUMERICALLY: 1. step: Find components of the resultant vector CC = AA + BB CC xx = AA xx + BB xx = AA cos θθ AA + BBBBBBBB θθ BB CC yy = AA yy + BB yy = AA ssssss θθ AA + BBBBBBBB θθ BB. step: Knowing components of the resultant vector, find its magnitude and direction: CC = CC xx + CC yy ; θ from the picture CC = CC(uuuuuuuuuu), θθ EXAMPLE: F = F = 3 65 Find F = F + F 1 F x = F 1x + F x = 68 cos cos65 0 = 75.6 N F y = F 1y + F y = 68 sin sin65 0 = 56.7 N F = F x + F y = 94.5N θ = arc tan (56.7/75.6) = 36.9 F 0 0 F =

3 3 DEFINITIONS and FORMULAS for 1-D KINEMATICS Displacement A change of position in particular direction. A distance in a given direction. Vector. Unit: meter (m) Average Velocity = displacement elapsed time v avg = x t Average Speed = distance traveled elapsed time v avg = d t (Instantaneous) Velocity 1. Value of velocity at a particular time. (Instantaneous) Speed 1. Value of speed at a particular time. Acceleration aa = vv cchaaaaaaaa iiii vvvvvvvvvvvvvvvv = tt tttttttt tttttttttt it has direction, the same as the change of velocity a = 3 m/s means that velocity changes 3 m/s every second!!! If an object s initial velocity is 4 m/s then after one second it will be 7 m/s, after two seconds 10 m/s, Acceleration can cause: 1. change in speed (speeding up: v nd a in the same direction; slowing down: v and a in the opposite direction). changing direction 3. both ALL TOGETHER: Any Motion Uniform Accelerated Motion equations ffffffff dddddddddddddddd: vv aaaaaa = xx tt xx = vv aaaaaa tt Motion with constant velocity uniform motion xx = vv aaaaaa tt aaaaaaaaaaaa ffffff aaaaaa mmmmmmmmmmmm v = v avg at all times, therefore: x = vt vv = uu + aaaa vv aaaaaa = uu + vv xx = uuuu + aa tt vv = uu + aaaa In addition to these equations to solve a problem with constant acceleration you ll need to introduce your own coordinate system, because displacement, velocity and acceleration are vectors (they have directions).

4 4 ALL WE KNOW ABOUT GRAPHS Velocity at some time is equal to the slope of the tangent line at that position on the position-time graph. Acceleration at some time is equal to the slope of the tangent line at that position on the velocity-time graph. Average velocity between two positions is equal to the slope of the secant line between these two points on the position-time graph. Average acceleration between two positions is equal to the slope of the secant line between these two points on the velocity-time graph. From velocity time graph : Area under the velocity-time graph between two times/position s is the displacement covered in that time interval. From acceleration time graph : Area under the acceleration-time graph between two times/positions is the change in velocity in that time interval.

5 5 Free Fall formulas Formulas are the ones for uniform accelerated motion with a = g v = u + gt v avg = u + v y = u + v t y = ut + g t v = u + gy g = 9.8 m/s, downward 10 m/s. Remember that in the coordinate system in which upward is chosen to be positive, g is negative and vice versa. If air resistance is not mention it is assumed that we ignore air resistance. When the object reaches maximum height, the velocity of the object is 0 m/s, but acceleration is still g = 9.8 m/s downward. Velocity changes, but g does NOT!!! Terminal speed/velocity is maximum velocity an object can reach in air/any fluid. Air resistance depends on velocity. The greater velocity, greater air resistance, smaller acceleration. Acceleration is getting smaller due to air resistance and eventually becomes zero when the force of the air resistance equals gravity, the downward force of gravity is equal to the upward force of air resistance resulting in a zero net force, hence zero acceleration. The object will stop accelerating and maintain the same speed. Terminal velocity/speed is different for different bodies. Graphs of the motion of an object in the gravitational field (downward is positive) PROBLEMS: 1. A motorist travels 400 km at 80 km/h and 400 km at 100 km/h. What is the average speed of the motorist on this trip? A. 84 km/h B. 89 km/h C. 90 km/h D. 91 km/h E. 95 km/h. A car starts from rest and uniformly accelerates to a final speed of 0.0 m/s in a time of 15.0 s. How far does the car travel during this time? A. 150 m B. 300 m C. 450 m D. 600 m E. 800 m 3. A car starts from rest and accelerates at 0.80 m/s for 10 s. It then continues at constant velocity. Twenty seconds after it began to move, the car has: A. velocity 8.0 m/s and has traveled 40 m.

6 6 B. velocity 8.0 m/s and has traveled 80 m. C. velocity 8.0 m/s and has traveled 10 m. D. velocity 16 m/s and has traveled 160 m. E. velocity 16 m/s and has traveled 30 m. 4. A racing car traveling with constant acceleration increases its speed from 10 m/s to 30 m/s over a distance of 80 m? How long does this take? 5. A rocket near the surface of the earth is accelerating vertically upward at 10 m/s. The rocket releases an instrument package. Immediately after release the acceleration of the instrument package is: A. 0 m/s up B. 10 m/s up C. 0 D. 10 m/s down E. 0 m/s down 6. A racing car traveling with constant acceleration increases its speed from 10 m/s to 30 m/s over a distance of 80 m? How long does this take? 7. An object with an initial velocity of 1 m/s west experiences a constant acceleration of 4 m/s west for 3 seconds. During this time the object travels a distance of: 8. An object starts from rest and accelerates uniformly in a straight line in the positive x direction. After 11 seconds, its speed is 70.0 m/s. a. Determine the acceleration of the object. b.how far does the object travel during the first 11 seconds? c.what is the average velocity of the object during the first 11 seconds? 9. A brick is dropped from rest from a height of 4.9 m. How long does it take for the brick to reach the ground? 10. What maximum height will be reached by a stone thrown straight up with an initial speed of 35 m/s? 11. An object dropped from a stationary balloon hits the ground in 1.0 s. If its acceleration is 9.80 m/s, the height of the balloon is: 1. How long does it take for a stone dropped off a 175-m high building to land on the ground? 13. A ball is thrown from the edge of 80 m high cliff, upward at speed of 30 m/s. At what two times are the displacements 10 m upward? What is velocity at these times? At what two times are the displacements 10 m downward? What is velocity at these times? 14. When a falling object reaches terminal velocity, it: A. is no longer subject to the friction of air. B. moves downward with constant velocity. C. has an acceleration of approximately 10 m/s. D. has no downward velocity. F. has an upward acceleration. 15. When a falling object moves with terminal velocity, it A. has zero velocity. B. has zero acceleration. C. has an upward acceleration.

7 7 D. is no longer subject to air resistance. E. has an acceleration of approximately 10 m/s. 1. B. A 3. C s 5. D s m 8. a.+6.4 m/s b. 390 m c. +35 m/s s m m 1. 6 s 14. B 15. B GRAPHS OF 1 D MOTION PROBLEMS 1. Given the graph of the velocity vs. time of a chick flying due south for the winter. At what B point did the duck stop its forward motion? A. A B. B C. C D. D E. none of these points. The graph represents tile relationship between distance and time for an object that is moving along a straight line. What is the instantaneous speed of the object at t = 5.0 seconds? A. 0.0 m/s B. 0.8 m/s C..5 m/s D. 4.0 m/s E. 6.8 m/s Between what times did the object have a non-zero acceleration? A. 0 s only B. 0 s to 5 s C. 5 s to 8 s B. 0 s to 8 s E. the object was not accelerating at any time. 3. The motion of a circus clown on a unicycle moving in a straight line is shown in the graph. a. What would be the acceleration of the clown at 5 s? A. 1.6 m!s B..0 m!s C. 3.4 m!s D) 8.O m!s E) none of the above b. After 1 seconds, how far is the clown from her original starting point? A. O m B. 10 m C. 34 m D. 47m E. 74m 4. The accompanying graph describes the motion of a marble on a table top for 10 seconds. a. For which time interval(s) did the marble have a negative velocity? A. from t = 8.0 s to t = 10.0 s only B. from t = 6.9 s to t = 10.0 s only C. from t = 4.8 s to t = 10.0 s only

8 8 D. from t = 4.8 s to t = 6. s and from t = 6.9 s to t = 10.O s only E. from t = 3. s to t = 3.6 s, from t = 4.8 s to t = 5 s, and from t = 6.8 s to t= 7. s only. b. For which time interval(s) did the marble have a positive acceleration? A. from t = 8.0 s to t = 8.0 s only B. from t = 0.0 s to t = 3.6 s only C. from t = 3.8 s to t = 4.8 s and from t = 6. s to t = 6.8 s only D. from t =.0 s to t =.5 s, from t = 5.8 s to t = 6. s, and from t = 8.4 s to t = 8.8 s only. E. from t = 3. s to t = 3.7 s, from t = 4.8 s to t = 5.0 s, and from t = 6.8 s to t= 7. s only. c. Vhat is the marbles average acceleration between t = 3.3 s and t = 3.8 s A. -.0 m/s B. 0.0 m/s C. 0.8 m/s D..0 m/s E. 3.0 m/s 5. The accompanying graph describes the motion of a toy car across the floor for 10 seconds. a. What is the acceleration of the toy car at t = 4 s? A. -1 m/s B. 0 m/s C. 1 m/s D. m/s E. 4 m/s b. What was the total displacement of the toy car for the entire 10 second interval shown? A. 0 meters B. 6.5 meters C. 9 meters D. 10 meters E meters 6. Graph refers to the motion of a toy car traveling along the x-axis. It is a plot of the car s velocity in the x direction. a. During what time interval was the car moving towards its initial position at constant velocity? A. 0-1Os B. 10-0s C. 0-5 s D s E s b. What was the acceleration at 33 s? A m/s B m/s C. 0 D m/s E m/s c. How far did the car travel during first 15 seconds? A. 0 B. 3.0 m C. 15 m D. 30 m E. 45 m 1. C. B, C 3. B, E 4. D, D, A 5. C, B 6. D, B, D

9 9 Uniform Circular Motion An object is moving in circular path at constant speed. Average velocity is zero, because displacement for a full circle is zero. velocity changing direction - tangent to the path (circle) and constant magnitude (speed) centripetal acceleration points toward the center and constant magnitude velocity and centripetal acceleration vectors are always perpendicular to each other speed v = distance/time v = ππππ r radius of the circle period T is time required for one complete revolution TT Centripetal acceleration that maintains motion around the circle with radius r, with speed v is given aa cc = vv rr ( aa cc ) = mm/ss PROBLEMS 1) A girl sits on a tire that is attached to an overhanging tree limb by a rope. The girl's father pushes her so that her centripetal acceleration is 3.0 m/s. If the length of the rope is.1 m, what is the girl's speed? ) A rock tied to a string is traveling at a constant speed of 4 m/s in a circle of radius 1.5 m. Calculate the magnitude of the centripetal acceleration of the rock. What is the direction of the acceleration? 3. A rock tied to the end of a string moves in a circle at a constant speed of.5 m/s and experiences an acceleration of 4.0 m/s. What is the radius of the circle of its motion? 1) v=.5m/s ) a c = 10.7 m/s 3) r = 1.56 m Projectile motion HORIZONTAL MOTION VERTICAL MOTION u x = u cos θ 0 u y = u sin θ 0 v x = u x v y = u y + gt x = u x t g y = uyt + t v =u + gy y y uy + vy y = t Problem: 1. A kid is playing with the pebbles at the top of a beautiful cliff above the sea. He flicks a pebble horizontally with the speed of 10 m/s. The pebble hits the sea 3s later. a. How high is the cliff above the sea? b. Find the speed and velocity s after it was thrown. Find the speed and velocity just before it hits the sea. c. Draw graphs: displacement vs. time, velocity vs. time, acceleration vs. time for each component of the motion. Write equation in each case.. In order to appreciate variation of speed and velocity during projectile motion, we calculate the values of a projectile velocity and speed for successive seconds t = 1s, 3s, and 5s, which is projected with an initial velocity of 60 m/s making an angle of 30 0 with the horizontal. Is that object on the way up or on the way down at those times. Find position (x,y) at the same times. How long does it take for that object to reach the same height? What is the maximum height? Draw graphs: displacement vs. time, velocity vs. time, acceleration vs. time for each component of the motion. Write equation in each case.