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

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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 v_average during during the time interval from 2 s to 2.1 s is given by

v_average = (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.