Question

A truck which was moving at a velocity of 20 m/s stopped after its brakes were applied. The distance from the instance the brakes were applied to the instance when it stopped was 490 m. Find:
(a) the acceleration of the truck
(b) the time taken to stop the truck

Answer 1

Step-by-step explanation:

Given:

Initial velocity of truck= 20m/s

Final velocity of truck= 0m/s ( as it stops)

Travelled distance= 490 m

By using 3rd equation of motion-

v²=u²+ 2as

0²= 20² + 2×a×490

0= 400 +2a× 490

2a= 0-400/490

2a = -40/49

a= (-40/49) / 2

a= - 0.408m/s² (after solving)

Time taken = t

Using 1st equation of motion

v = u + at

Then,

t = v - u /a

t = 0 - 20 / - 0.408

t = 49.01 sec ( after solving)