Question

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

Answer 1

Answer:

(A) Original velocity will be 7.244 m /sec

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

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

$S=ut+\frac{1}{2}at^2$

$40=u\times 7.45+\frac{1}{2}\times a\times 7.45^2$

$7.45u+27.751a=40$-----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

$-0.5026m/sec^2$