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AP Physics C Unit 1 Review Problems. North Allegheny Senior High Kernion. Unit Vectors and Uniform Circular motion. An object moves in uniform circular motion. At t = 2 sec it has a velocity (in m/s) of 3 i + 4 j . at t = 5 sec it has a velocity

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AP Physics C Unit 1 ReviewProblems North Allegheny Senior High Kernion

Unit Vectors and Uniform Circular motion An object moves in uniform circular motion. At t = 2 sec it has a velocity (in m/s) of 3i + 4 j. at t = 5 sec it has a velocity (in m/s) of -3i – 4 j. Draw a sketch of this motion and find the radius.

Unit Vectors and Uniform Circular motion An object moves in uniform circular motion. At t = 2 sec it has a velocity (in m/s) of 3i + 4 j. at t = 5 sec it has a velocity (in m/s) of -3i – 4 j. Draw a sketch of this motion and find the radius. The points shown represent motion that is half-way around the trajectory. Thus, T = 6 seconds -3i -4j 4j 3i

Unit Vectors and Uniform Circular motion An object moves in uniform circular motion. At t = 2 sec it has a velocity (in m/s) of 3i + 4 j. at t = 5 sec it has a velocity (in m/s) of -3i – 4 j. Draw a sketch of this motion and find the radius. Note: The average acceleration is NOT The same as the instantaneous acceleration v = [(32) + (4)2]1/2 v = 5 m/s 2pr / T = v T = 2pr/v 6 = 2pr/5r = 4.4 m a = 5.7 Now, since ainst = v2/r ainst = (5)2 /4.4 = 5.7 m/s2 This acceleration magnitude is the SAME at every point. But the direction is always changing. a = 5.7 a = 5.7

Unit Vectors and Uniform Circular motion An object moves in uniform circular motion. At t = 2 sec it has a velocity (in m/s) of 3i + 4 j. at t = 5 sec it has a velocity (in m/s) of -3i – 4 j. Draw a sketch of this motion and find the radius. Note: The average acceleration is NOT The same as the instantaneous acceleration because we are not moving along one axis and in one direction. For example: If a car were uniformly accelerating west for an entire experiment then the average and the instantaneous accelerations would be the same the entire trip. BUT…If a car uniformly accelerated west with a certain magnitude and then uniformly accelerated east for the exact same amount of time, the average acceleration would be zero and the instantaneous acceleration would have the same MAGNITUDE the entire time

Unit Vectors and Uniform Circular motion An object moves in uniform circular motion. At t = 2 sec it has a velocity (in m/s) of 3i + 4 j. at t = 5 sec it has a velocity (in m/s) of -3i – 4 j. Draw a sketch of this motion and find the radius. The same type of thing is true for the uniform circular motion case… Because the acceleration is not uniformly in the same direction, the average acceleration vector and the instantaneous acceleration vectors DO NOT have the same magnitudes, even though the magnitude of the instantaneous acceleration is uniform over the entire trajectory. aave= Dv/Dt = (-3i – 4j) – (3i + 4j) / 3 aave = (-6i – 8j) / 3 = -2i – 2.7j One other way to look at it: The average acceleration over one complete cycle is ZERO even though the instantaneous acceleration at each point is 5.7 m/s2 (magnitude) average acceleration magnitude aave = [(2)2 + (2.7)2)]1/2 = 3.4 m/s2

Unit Vectors and Kinematics Challenge The velocity of a particle moving in the x-y plane is given by: v = ( 6.0 t – 4.0 t2 )i + 8.0 j 1. What is the acceleration when t = 3.0 seconds? 2. When, if ever, is the acceleration zero? 3. When, if ever, is the speed zero? 4. When, if ever, does the speed equal 10.0 m/s?

Solutions v = ( 6.0 t – 4.0 t2 )i + 8.0 j 1. What is the acceleration when t = 3.0 seconds? a = dv/dt = [d( 6.0 t – 4.0 t2 )i + 8.0 j)]/dt a = (6 – 8t) i a(3.0) = (6 – 24) i = -18 i m/s2

Solutions v = ( 6.0 t – 4.0 t2 )i + 8.0 j a = (6 – 8t) i 2. When, if ever is a = 0? a = (6 – 8t) i = 0 when 6 – 8t =0 6 = 8t t = 0.75 s

Solutions 3. When, if ever, does the speed equal zero? v = (6t – 4 t2) i + 8 j since the speed in the y-direction is constant, the overall speed never equals zero.

Solutions When, if ever, does the speed equal 10 m/s? v = (6t – 4t2) i + 8 j v = [vx2 + vy2]1/2 v = [(6t – 4t2)2 + 82 ]1/2 = 10 (6t – 4t2)2 + 82 = 100 (6t – 4t2)2 = 36 – 4t2 + 6t (+/-) 6 = 0 (quadratic) t = 2.2 s The square root of 36 is + 6 or – 6

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