Question

A truck is going 30 m/s when the driver applies the brakes for 4.7 s so that the truck slows down to 20 m/s. a) What is the acceleration of the truck? b) How far does the truck go during the 4.7 s?

Answer 1

(a) The acceleration of the truck is,
`-2.1m/s^2`

(b) The distance covered by the truck is, 117.8 m

Explanation :

By the 1st equation of motion,

`v=u+at` ...........(1)

where,

v = final velocity = 20 m/s

u = initial velocity = 30 m/s

t = time = 4.7 s

a = acceleration of the truck = ?

Now put all the given values in the above equation 1, we get:

`20m/s=30m/s+a* (4.7s)`

`a=-2.1m/s^2`

The acceleration of the truck is,
`-2.1m/s^2`

By the 2nd equation of motion,

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

where,

s = distance covered by the truck = ?

u = initial velocity = 30 m/s

t = time = 4.7 s

a = acceleration of the truck =
`-2.1m/s^2`

Now put all the given values in the above equation 2, we get:

`s=(30m/s)* (4.7s)+(1)/(2)* (-2.1m/s^2)* (4.7s)^2`

By solving the term, we get:

`s=117.8m`

The distance covered by the truck is, 117.8 m