Question

A car moves linearly with uniform retardation. If the car covers 40m in the last 2 seconds of its motion, what is the velocity of the car at the beginning of the last second

Answer 1

First, let us find the relation between the constant retardation a and the initial velocity u at the beginning of the last 2 second.

we know that,

V2−U2=2aSV^{2} -U^{2} =2aSV2−U2=2aS

Substituting final velocity V=0 and displacement S=40 m we get

=>−U2=80(−a)=> -U^{2}=80(-a)=>−U2=80(−a)

a=U2/80a= U^{2}/80a=U2/80...........1

Keeping that aside,

Apply

$v=u+at$

And substituting value of a from equation 1

V=U+(−U2/80)2V=U+ (-U^{2}/80)2V=U+(−U2/80)2

Since we know that V=0

0=U+(−U2/80)20=U+ (-U^{2}/80)20=U+(−U2/80)2

U=U2/40U=U^{2}/40U=U2/40

40U=U240U=U^{2}40U=U2

$U=40m/s$

.’. The initial velocity U=40 m/s

This is the velocity of the car at the beginning of the last 2 seconds.

According to the question we have to find the velocity at the beginning of the last second

Now,

Initial velocity = 40 m/s

Acceleration = $\frac{40\times 40}{80}=-20$ m/$s^2$

Time = 1 sec

According to the equation of motion

$v=u+at$

Putting all the values

$v=40-20\times 1=20$ m/s

Hence, the velocity of the car at the beginning of the last seconds is 20 m/s