“God used beautiful mathematics in creating
the world.”

– P.A.M. Dirac

Today we shall discuss a few questions (MCQ) in
the section, ‘kinematics in one dimension’. The questions I give you are meant
for testing your knowledge, comprehension and the ability for applying what you
have learned in this section.

(1) The velocity-time graphs of a car and a motor
bike traveling along a straight road are shown in the adjoining figure. At time

*t*= 0 they have the same position co-ordinate. Pick out the correct statement from the following:
(a) At time

*t =*0 the car and the motor bike are at rest.
(b) At time

*t =*0 the car is at rest but the motor bike is moving.
(c) The distances traveled by the car and the
motor bike in time

*t*_{1}are equal.
(d) The distance traveled by the bike in time

*t*_{1}is twice the distance traveled by the car in the same time.
(e) The distance traveled by the bike in time

*t*_{1}is half the distance traveled by the car in the same time.
You can easily conclude that options (a) and (b)
are wrong.

To check the remaining options we use the
equation of linear motion,

*s*=*ut +*½*at*^{2}where*s*is the displacement in time*t*,*u*is the initial velocity*and**a*is the uniform acceleration.
The distance traveled by the car in time

*t*_{1}is*v*_{c}*t*_{1}where*v*_{c}(let us say) is the constant velocity of the car.
[The velocity of the car is constant since the
velocity-time graph of the car is parallel to the time axis].

The motion of the motor bike is uniformly
accelerated and the acceleration

*a*is given by*a =*

*v*

_{c}/t

_{1}

[Note that the velocity of the motor bike changes
from 0 to

*v*_{c}in time*t*_{1}.
The distance

*s*traveled by the motor bike in time*t*_{1 }is given by*s*=

*ut +*½

*at*

^{2}= 0 + (½)×(

*v*

_{c}/t

_{1})

*t*

_{1}

^{2}

Or,

*s*=*v*_{c}*t*_{1}/2
Therefore the distance traveled by the bike in time

*t*_{1}is half the distance traveled by the car in the same time [Option (e)].
(2) A student standing at the edge of a cliff
throws a stone of mass

*m*vertically upwards with speed*v*. It strikes the ground at the foot of the cliff with speed*v*_{1}. The student then throws another stone of mass*m/*4 vertically downwards with speed*v*. It strikes the ground at the foot of the cliff with speed*v*_{2}. If air resistance is negligible,*v*_{1}and*v*_{2}are related as
(a)

*v*_{1}=*v*_{2}
(b)

*v*_{1}=*v*_{2}/2
(c)

*v*_{1}= 2*v*_{2}
(d)

*v*_{1}=*v*_{2}/4
(e)

*v*_{1}= 4*v*_{2}
While returning, the sphere of mass

*m*has*downward*speed*v*when it passes the edge of the cliff. For calculating the speed with which it strikes the ground at the foot of the cliff, the situation is similar to that of the stone of mass*m/*4 thrown downwards. Obviously both stones will strike the ground with the*same*speed [Option (a)].
(3) A body is projected vertically upwards. The
times corresponding to height

*h*while ascending and descending are*t*_{1}and*t*_{2}are respectively. Then the velocity of projection is (*g*is acceleration due to gravity)
(1)

*g*√(*t*_{1}*t*_{2})
(2)

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

*g*√(*t*_{1}*t*_{2})/2
(4)

*g*(*t*_{1}+*t*_{2})/2
The above question appeared in Karnataka CET 2008
question paper.

We have the following two equations to give

*h*:*h = ut*

_{1}– (½)

*g t*

_{1}

^{2}…………….(i)

*h = ut*

_{2}– (½)

*g t*

_{2}

^{2}…………….(ii)

(We have taken the displacement

*h*and the initial velocity*u*(both*upwards*) as*positive*and that’s why the acceleration due to gravity*g*is negative).
(ii) – (i)
gives

*u*(*t*_{2}–*t*_{1}) = (½)*g*(*t*_{2}^{2 }–*t*_{1}^{2})
Therefore

*u*= (½)*g*(*t*_{2}^{2 }–*t*_{1}^{2})/(*t*_{2}–*t*_{1})*= g*(*t*_{1}+*t*_{2})/2
You can find a few more multiple choice practice
questions (with solution) in this section here.

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