Question

A truck covers 40.0 m in 7.50 s while uniformly slowing down to a final velocity of 2.55 m/s. (a) Find the truck's original speed. (b) Find its acceleration.

Answer 1

(a) 8.117 (b) 0.742
`m/sec^2`

Step-by-step explanation:

We have given distance s =40 m time t=7.5 sec

Final velocity v =2.55 m/sec

From the first equation of motion
`v=u-at` (negative sign because there is retrdation as the truck speed is slowing down )

So
`2.55=u-7.5a` --------------eqn 1

From the second equation of motion
`s=ut-(1)/(2)at^2` ( negative sign because there is retrdation as the truck speed is slowing down )

So
`40=7.5u-0.5* a* 7.5^2`

`40=7.5u-28.125a`------------------eqn 2

On solving eqn1 and eqn 2

u=8.117 m/sec and a=-0.742
`m/sec^2`