Atruck 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.

(a) 8.117 (b) 0.742

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 (negative sign because there is retrdation as the truck speed is slowing down )

So --------------eqn 1

From the second equation of motion ( negative sign because there is retrdation as the truck speed is slowing down )

So

------------------eqn 2

On solving eqn1 and eqn 2

u=8.117 m/sec and a=-0.742

the second one and the forth one are in the purple because they are opinions and the rest are in the yellow

explanation:

v=ir = 3x15 =45 v