“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

*gt*^{2}
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

*s*_{1}traveled by the object in the 1^{st}second is given by*s*

_{1}= (50×1) – (½ ×10×1

^{2}) = 45 m

[We have used the equation,

*s = ut +*½*at*^{2}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*s*_{6}*traveled by the object during the next one second (6*^{th}second) is given by*s*

_{6}= 0 + ½ ×10×1

^{2}= 5 m.

[We have used the
equation of motion,

*s**= ut +*½*gt*^{2}with usual notations]
The ratio of the distance
traveled by the object in the 1

^{st}second and the 6^{th}second is*s*

_{1}:

*s*

_{6}= 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) 1.75

*t*
(b) 1.65

*t*
(c) 1.5

*t*
(d) 1.4

*t*
(e) 1.3

*t*
From the equation of motion,

*v*^{2}=*u*^{2}+ 2*as*with usual notations, we have for the upward journey
0
=

*u*^{2}– 2*gh*
Therefore

*h*=*u*^{2}/2*g*
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*=

*gt*

^{2}/2 …………. (i)

We now use the equation of motion,

*s = ut +*½*gt*^{2}for the free fall of the ball from the maximum height*h*to the height*h/*4.
Since distance of fall is 3

*h*/4) we have
3

*h*/4 = 0*+*½*gt*_{1}^{2}where*t*_{1}is the time of fall from the maximum height*h*to the height*h/*4.
Substituting for

*h*from equation (i) we have*gt*

^{2}/8 =

*gt*

_{1}

^{2}/2

This gives

*t*_{1}=_{ }*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 +**t*_{1}=*t +*(*t/*_{2}) = 3*t/*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*gt*^{2}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)×1×2

^{2}= 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 =*6*t*,*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*= ∫6

*t dt*= 3

*t*

^{2}+

*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*= 3

*t*

^{2}+ 2

The position

*x*of the particle is given by*x =*∫

*v dt =*∫(3

*t*

^{2}+ 2)

*dt = t*

^{3}+ 2

*t + 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 = t*

^{3}+ 2

*t*

The position

*x’*at*t =*4 seconds is given by*x’ =*4

^{3}+ (2×4) = 72 m

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

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