Question

A truck covers 40.0 m in 7.45 s while uniformly slowing down to a final velocity of 3.50 m/s.

(a) Find its original speed.
(b) Find its acceleration

Answer 1

(A) Original velocity will be 7.244 m /sec

(B) Acceleration will be
`-0.5026m/sec^2`

Step-by-step explanation:

We have given distance covers by truck s = 40 m

Time taken by truck to cover this distance t = 7.45 sec

Final velocity v = 3.50 sec

According to second equation of motion

`S=ut+(1)/(2)at^2`

`40=u* 7.45+(1)/(2)* a* 7.45^2`

`7.45u+27.751a=40`-----eqn 1

According to first equation of motion

v = u + at

So
`3.5=u+7.45a`-----eqn2

Solving equation 1 and 2

a =
`-0.5026m/sec^2`

And u = 7.244 m /sec

(A) Original velocity will be 7.244 m /sec

(B) Acceleration will be

`-0.5026m/sec^2`