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## Thursday, November 19, 2009

### Kinematics - Practice Questions (MCQ) for AP Physics B & C

“Clarity about the aims and problems of socialism is of greatest significance in our age of transition.”

Albert Einstein

You can access all posts on kinematics on this site by clicking on the label ‘kinematics’ below this post. Or, you may try searching for ‘kinematics’ using the search box at the top of this page.
Today I give you three more multiple choice practice questions with solution:
(1) Tom runs from his school to his home with uniform speed v1 and returns to his school with uniform speed v2. The average speed of his round trip is
(a) (v1 + v2)/2
(b) [(v12 + v22)/2]1/2
(c) (v1v2)1/2
(d) v1v2/(v1 + v2)
(e) 2v1v2/(v1 + v2)
The average speed is the ratio of the total distance traveled to the total time taken.
If the distance between Tom’s school and his home is s, the total distance for the round trip is 2s and the total time for the round trip is (s/v1 + s/v2).
Therefore, average speed = 2s/(s/v1 + s/v2) = 2v1v2/(v1 + v2)
(2) The displacement ‘y’ of a particle thrown vertically down is given by the equation, y = 2t + 5t2, where y is in metre and t is in second. The average velocity during the time interval from 2 s to 2.1 s is
(a) 12.5 ms–1
(b) 16 ms–1
(c) 20.5 ms–1
(d) 22.5 ms–1
(e) 32 ms–1
The velocity ‘v’ of the particle at the instant t is given by
v = dy/dt = 2 + 10t
Therefore, velocities at instants 2 s and 2.1 s are respectively 2 + 10×2 = 22 ms–1 and 2 + 10×2.1 = 23 ms–1.
The average velocity vaverage during during the time interval from 2 s to 2.1 s is given by
vaverage = (22 + 23)/2 = 22.5 ms–1
[The above question can be answered without using calculus (as in the case of some of the AP Physics B aspirants) like this:
The displacement y is in the form for uniformly accelerated motion in one dimension: y = v0t + ½ at2 where v0 is the initial velocity and ‘a’ is the acceleration. Therefore, v0 = 2 ms–1 and a = 10 ms–2.
The velocity ‘v’ at the instant t is given by v = v0 + at = 2 + 10t].
The following question is meant specifically for AP Physics C aspirants:
(3) At the instant t, the position ‘x’ of a particle moving along the x-axis is given by x = 12t2 – 2t3 where x is in metre and t is in second. What will be the position of this particle when it moves with the maximum speed along the positive x direction?
(a) 32 m
(b) 36 m
(c) 40 m
(d) 48 m
(e) 52 m
The speed v of the particle is given by
v = dx/dt = 24t – 6t2.
When the speed is maximum we have dv/dt = 0
Therefore, 24 – 12t = 0 from which t = 2 s.
The position of the particle at 2 seconds will be 12×(2)2 – 2×(2)3 = 48 – 16 = 32 m (as obtained from x = 12t2 – 2t3).

Useful posts in this section can be seen at physicsplus.