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

Answer :

(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:

$20m/s=30m/s+a\times (4.7s)$

$a=-2.1m/s^2$

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

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(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:

$s=(30m/s)\times (4.7s)+\frac{1}{2}\times (-2.1m/s^2)\times (4.7s)^2$

By solving the term, we get:

$s=117.8m$

The distance covered by the truck is, 117.8 m