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

Answer:

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

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 $s=ut-\frac{1}{2}at^2$ ( negative sign because there is retrdation as the truck speed is slowing down )

So $40=7.5u-0.5\times a\times 7.5^2$

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

On solving eqn1 and eqn 2

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