Question

A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.

Answer 1

Step-by-step explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

`a=(v-u)/(t)`.............(1)

Now, the second equation of motion is :

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

Put the value of a in above equation as :

`s=ut+(1)/(2)* (v-u)/(t)* t^2`

`s=(t(u+v))/(2)`

`u=(2s)/(t)-v`

`u=(2* 47)/(8.6)-2.3`

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.