AP Physics B and C – Kinematics – Answers to Multiple Choice Questions on One Dimensional Motion

Some typical multiple choice questions (for practice) on one dimensional motion were given to you in the post dated 18^th August 2008. As promised, I give below the answers with explanation along with those questions:

Questions for AP Physics B and C:

(1) The adjoining figure shows the displacement-time graph of a particle in uniformly accelerated linear motion. At the points A and B (fig.) the graph is inclined at 30º and 60º respectively with the time axis. The acceleration of the particle in ms^–2 is

(a) √2

(b) √3

(c) 3/√2

(d) 2/√3

(e) 1/√3

The slope of the displacement-time graph gives the velocity and hence the velocity of the particle at time 2 s is tan30º = 1/√3. Similarly, the velocity of the particle at time 3 s is tan 60º = √3.

Acceleration = change of velocity/time = [√3 – (1/√3)]/(3 – 2) = 2/√3.

(2) A stone released from the top of a tower falls through half the height of the tower in 3/√2 seconds. The total time taken by the stone to reach the ground is

(a) 3 s

(b) 3.3 s

(c) 3.5 s

(d) 3.8 s

(e) 4 s

If h is the height of the tower we have

h/2 = 0 + (½) g (3/√2)²

This gives h = 9g/2

If t is the total time taken to reach the ground, we have

9g/2 = 0 + (½) g t² so that t = 3 s.

(3) A particle in one dimensional motion has initial velocity of 1 ms^–1. The variation of its acceleration with time is represented by the semi circular curve shown in the adjoining figure. What is the velocity of the particle ( in ms^–1) at the end of 8 seconds?

(a) 2π

(b) 2π + 1

(c) 4π

(d) 4π + 1

(e) 1

The area under the acceleration-time graph gives the change in velocity. [This area gives the sum of the products of the acceleration and time at all instants indicated by the curve. Since acceleration = change in velocity/time, the products of acceleration and time is the change in velocity and the area under the curve indeed gives the total change in the velocity].

Therefore, the change in velocity during 4 s = Area of the semicircle.

= (½)π ×4×2 = 4π

[Since the x-plot and the y-plot are of different quantities and the scales can be chosen arbitrarily, you have to treat the shape as half ellipse and the area is (½)πab where a and b are the semi major axis and semi minor axis respectively]

Since the initial velocity is 1 ms^–1, the velocity at the end of 4 s = (4π +1) ms^–1

(4) The bob of a simple pendulum is made to move along a semicircular path of radius 1 m. If the time taken to move along the semicircle is 1 s, the average velocity of the bob is

(a) π ms^–1

(b) π/2 ms^–1

(c) 1 ms^–1

(d) 2 ms^–1

(e) zero

The displacement of the bob in 1 s is 2 m.

Therefore, the average velocity of the bob = Displacement/Time = 2 m/1 s = 2 ms^–1.

(5) A stone is projected vertically upwards from point A (fig.) at the foot of a tower. It passes the top level B of the tower after 2 s. After another 2 s it again reaches the top level of the tower while returning from the top point C of its path. What is the height of the tower?

(a) 98 m

(b) 55 m

(c) 40 m

(d) 22 m

(e) 10 m

The time required for the upward journey of the stone is equal to the time required for the downward journey.

Time required for the upward journey (from A to C) = 2 + (2/2) = 3 s.

You can imagine that a stone released from rest at the point C takes a time of 3 s for the fall from C to A so that distance CA = (½)gt² = (½)×10×3² = 45 m.

Similarly, CB = (½)×10×1² = 5 m.

Therefore, height of the tower, BA = 45 – 5 = 40 m.

Questions for AP Physics C:

(1) The displacement x of a particle moving along a straight line is given by

x = 3t² + 2t + 4 where x is in metre and t is in second.

If the particle could move for 3 seconds, what was its acceleration (in ms^–2) at t = 2 s?

(a) 15

(b) 6

(c) 4

(d) 3

(e) 2

The acceleration a = d²x/dt² = 6 ms^–2

Since this is independent of time t, the particle moves with uniform acceleration throughout the time interval. The correct option is 6 ms^–2.

(2) A particle moving along the X-direction has its displacement x related to time t as

t = α x² + β x where α and β are positive constants.

If the velocity of the particle at the instant t is v, its acceleration is

(a) α v²

(b) – α v²

(c) – 2 α v²

(d) 2α v³

(e) – 2α v³

We have t = α x² + β x. Differentiating with respect to t, we obtain

1 = 2αx dx/dt + β dx/dt

Therefore, dx/dt = 1/(2αx+ β)

Acceleration, a = d²x/dt² = – (2α dx/dt) /(2αx+ β)²

Since dx/dt = v, the above relation becomes a = – 2αv³

(3) A body falling freely under gravity covers half its total path in the last second of its fall. The total time taken by the body to cover the entire path is approximately

(a) 1.414 s

(b) 1.732 s

(c) 2.414 s

(d) 3.414 s

(e) 3.732 s

If h is the height from which the body falls and t is the total time of fall, we have

h = 0 + (½)gt² = 5t²

Also, h/2 = 0 + g(t – ½) = 10t – 5

Substituting for h from the first equation, 5t²/2 = 10t – 5

Or, t² – 4t + 2 = 0

This gives t = 2 ±√2

Since t has to be greater than 1 s, as indicated in the question (there is a last second for the fall!), the answer is (2 + √2) s = 3.414 s.

(4) The co-ordinates of a moving particle at time t are given by x = a_x t², y = a_y t² and z = a_z t² where a_x, a_y, and a_z are constants. The magnitude of its velocity at time t is

(a) 2t (a_x² + a_y² + a_z²)^1/2

(b) t (a_x² + a_y² + a_z²)^1/2

(c) 2t (a_x² + a_y² + a_z²)

(d) t (a_x² + a_y² + a_z²)

(e) 2t (a_x + a_y + a_z)/3

The components of its velocity are

v_x = dx/dt = 2a_x t

v_y = dy/dt = 2a_y t

v_z = dz/dt = 2a_z t

The magnitude of the velocity is (v_x² + v_y² + v_z²)^1/2 = 2t (a_x² + a_y² + a_z²)^1/2

You teach me baseball and I'll teach you relativity... No, we must not. You will learn about relativity faster than I learn baseball.

– Albert Einstein