AP Physics C - Multiple Choice Practice Questions on One Dimensional Motion

“Maturity is often more absurd than youth and very frequently is most unjust to youth.”

– Thomas A. Edison

Today’s post covers a few multiple choice practice questions related to one dimensional kinematics meant for AP Physics C aspirants.

(1) Water drops from a leaking overhead tank falls on the ground 5 m below, at regular intervals, the 11^th  drop just beginning to fall when the first drop strikes the ground. What will be the height of the 9^th drop when the first drop strikes the ground? (Acceleration due to gravity = 10 ms^–2)

(a) 4.9 m

(b) 4.8 m

(c) 4.6 m

(d) 4.4 m

(e) 4.2 m

The time t taken by a drop to fall through 5 m is given by the relevant equation of motion,

            5 = 0 + ½ gt²

Since g = 10 ms^–2 the above equation gives t = 1 second

When the first drop strikes the ground after falling freely for one second, the 9^th  drop has fallen freely for one-fifth of one second, as is evident from the adjoining figure (remembering that the drops fall at regular intervals of time). Therefore, the 9^th drop has fallen through a distance s given by

            s = 0 + ½ g(0.2)² = (½)×10×0.04 = 0.2 m

Therefore the height of the 9^th drop when the first drop strikes the ground is 5 m – 0.2 m = 4.8 m.

(2) An object projected vertically upwards with initial velocity u attains maximum height in 5 s. The ratio of the distance traveled by the object in the 1^st second and the 6^th second is (Acceleration due to gravity = 10 ms^–2)

(a) 6 : 1

(b) 8 : 1

(c) 9 : 1

(d) 10: 1

(e) 11 : 1

The velocity of projection (u) of the object is given by the equation of uniformly accelerated linear motion,

             0 = u – gt

[We have used the equation, v = u + at with usual notations].

Substituting for g and t we have

             0 = u – 10×5

Therefore u = 50 ms^–1

The distance s₁ traveled by the object in the 1^st second is given by

             s₁ = (50×1) – (½ ×10×1²) = 45 m

[We have used the equation, s = ut + ½ at² with usual notations]

At the end of 5 seconds the object is at the highest point of its trajectory where its velocity is zero. Therefore, the distance s₆ traveled by the object during the next one second (6^th second) is given by

             s₆ = 0 + ½ ×10×1² = 5 m.

[We have used the equation of motion, s = ut + ½ gt² with usual notations]

The ratio of the distance traveled by the object in the 1^st second and the 6^th second is

             s₁ : s₆ = 45 : 5 = 9 : 1

(3) A ball projected vertically upwards with initial velocity u reaches maximum height h in t sec. What is the total time taken by the ball (from the instant of projection) to reach a height h/4 while returning?

(a) 1.75t

(b) 1.65t

(c) 1.5t

(d) 1.4t

(e) 1.3t

From the equation of motion, v² = u² + 2as with usual notations, we have for the upward journey

             0 = u² – 2gh

Therefore h = u²/2g

From the equation of motion, v = u + at with usual notations, we have for the upward journey

             0 = u – gt so that u = gt

Substituting for u in the expression for h we have

             h = gt²/2 …………. (i)

We now use the equation of motion, s = ut + ½ gt² for the free fall of the ball from the maximum height h to the height h/4.

Since distance of fall is 3h/4) we have

             3h/4 = 0 + ½ gt₁² where t₁ is the time of fall from the maximum height h to the height h/4.

Substituting for h from equation (i) we have

             gt²/8 = gt₁²/2

This gives t₁ = t/2

The total time time taken by the ball (from the instant of projection) to reach a height h/4 while returning is t + t₁ = t + (t/₂) = 3t/2 = 1.5 t.

(4) The velocity-time graph of an object moving along the x-direction is shown in the figure. What is the displacement of the object when it moves with the maximum acceleration?

(a) 12 m

(b) 8 m

(c) 6 m

(d) 4.8 m

(e) 4.2 m

The object has maximum acceleration from 10 sec to 12 sec since the slope of the velocity-time graph is maximum during this interval. The displacement is given by the area under the velocity-time graph for the interval from 10 sec to 12 sec which is 8 m.

[The displacement of the object when it moves with the maximum acceleration can be calculated using the equation,  s = ut + ½ gt² as well:

As is evident from the velocity-time graph, the maximum acceleration a is given by

             a = Change of velocity/Time = (5 – 3)/(12 – 10) = 1 ms^–2

Since the initial velocity u = 3 ms^–1 and the time interval t = 2 s, we have

             s = (3×2) + (½)×1×2² = 8 m].

(5) A particle moving along the x-axis is initially at the origin with velocity 2 ms^–1. If the acceleration a of the particle is given by a = 6t, the position of the particle after 4 seconds is

(a) 24 m

(b) 48 m

(c) 72 m

(d) 96 m

(e) 120 m

The velocity v of the particle is given by

             v = ∫adt = ∫6t dt = 3t² + C where C is the constant of integration which we can find from the initial condition.

Initially (at time t = 0) the particle has velocity 2 ms^–1. Therefore from the above expression for velocity we have

             C = 2 ms^–1

Thus the expression for velocity becomes

             v = 3t² + 2

The position x of the particle is given by

             x = ∫v dt = ∫(3t² + 2)dt = t³ + 2t + C’ where C’ is the constant of integration in this case.

Initially (at t = 0) since the particle is at the origin (where x = 0) we obtain (from the above expression for v)

             C’ = 0

Therefore, the expression for the position x becomes

             x = t³ + 2t

The position x’ at t = 4 seconds is given by

             x’ = 4³ + (2×4) = 72 m

You can find a few more useful questions (with solution) in this section here.