"Whenever you are confronted with an
opponent, conquer him with love."

– Mahatma Gandhi

Let us discuss a few interesting multiple choice practice questions on one dimensional motion. Here are some questions beneficial for AP Physics B as well as AP Physics C aspirants:

(1) The adjoining
figure shows the velocity time graph of an object. Total displacement suffered
by the object during the interval when it has non-zero acceleration and
retardation is

(a) 80 m

(b) 70 m

(c) 60 m

(d) 40 m

(e) 30 m

The object has non-zero
acceleration and retardation during the time intervals from 5 sec to 15 sec and
from 20 sec to 30 sec. The total area under the velocity time graph during
these intervals gives the required displacement.

Displacement from 5
sec to 15 sec = 30 m

Displacement from
20 sec to 30 sec = 40 m

Therefore, total
displacement = 70 m

(2) Successive
positions (

*x*) of an object (moving from left to right) at equal time intervals are shown in the following figure:
Which one among the
following position-time graphs best represents the motion of the object?

Change of position
is slowest in the beginning and in the end. The motion is best represented by
graph (c).

[Note that graph (d)
is not the answer since the change of position in the beginning and in the end
is shown as fastest in it]

(3) Which one among
the following velocity-time graphs best represents the motion of the object
mentioned in question no.(2)?

The graph (c) is
the answer.

*The following questions are meant for*

**AP Physics C**aspirants:
(4) A particle projected
vertically upwards attains the maximum height

*h*in time*t*. While returning from the highest point it takes an additional time*t*_{1}to fall to the height*h*.2. If air resistance is negligible, how is*t*_{1}related to*t*?
(a)

*t*_{1}=*t*/2
(b)

*t*_{1}= √(*t*/2)
(c)

*t*_{1}=*t*/√2
(d)

*t*_{1}=*t*/√3
(e)

*t*_{1}=*t*/3
For the upward
motion we have

0 –

*u*^{2}= – 2*gh*…………(i)
[We have used the
equation of motion,

*v*^{2 }–*u*^{2}= 2*as*.*The sign of the gravitational acceleration**g*is*negative*since it is opposite to the direction of the velocity of projection*u*]
Using the equation
of motion,

*v = u + at*we have
0 =

*u*–*gt*from which*u = gt*
Substituting this
value of

*u*in Eq (i), we have*g*

^{2}

*t*

^{2}= 2

*gh*

Therefore,

*t =*√(2*h*/*g*)*……..(ii)*
For the fall
through

*h*/2 from the highest point we have*h*/2 = 0×

*t*

_{1}+ ½

*gt*

_{1}

^{2}

[We have used the
equation of motion,

*s = ut +*½*gt*^{2}]
Therefore,

*t*_{1}*=*√(*h*/*g*)
Comparing this with
the value of

*t*given in Eq (ii) we obtain*t*_{1}=*t*/√2.
(5) A ball projected
vertically up has the same vertical displacement

*h*at times*t*_{1}second and*t*_{2}second. If air resistance is negligible, the maximum height reached by the ball is
(a)

*g**t*_{1}*t*_{2}/2
(b)

*g*(*t*_{1}^{2}+*t*_{2}^{2})/2
(c)

*g*(*t*_{1}+*t*_{2})^{2}/4
(d)

*g*(*t*_{1}+*t*_{2})^{2}/8
(e)

*2**g*(*t*_{1}+*t*_{2})^{2}
The time taken by
the ball to move from height

*h*to the top of its trajectory and back to the height*h*is*t*_{2}–*t*_{1}. Therefore, the time taken to move from height*h*to the top of the trajectory is (*t*_{2}–*t*_{1})/2.
The total time taken for the

*upward*journey (from ground to the top most point) is evidently*t*_{1}+*t*_{2}–*t*_{1})/2 = (*t*_{1}+*t*_{2})/2.
Time taken for the return journey (from the top most point
to the ground) also is equal to (

*t*_{1}+*t*_{2})/2. Therefore, the maximum height*H*(using the equation,*s = ut +*½*gt*^{2}) is given by*H =*0 + ½

*g*[(

*t*

_{1}+

*t*

_{2})/2]

^{2}

Or,

*H =**g*(*t*_{1}+*t*_{2})^{2}/8
Questions on kinematics were discussed earlier on
this site. You can access them either by clicking on the label ‘kinematics’
below this post or by trying a search for ‘kinematics’ using
the search box provided on this page.

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