Question

a car moving at a rate of 72km/h applies brakes which provide a retardation of 5m/s*s.
i)how much time does the car takes to stop?
ii)how much distance does the car cover before coming to rest?
iii)what would be the stopping distance needed if speed of the car is doubled?

Answer 1

Initial velocity(u) = 72 km/h = 20m/s

Final velocity(v) = 0 (as it comes to rest)

Acceleration = -5m/s2 (It is negative 5m/s2 as it is retardation)

 i) Time =?

1st equation o motion (v = u +at)

0 = 20m/s + (-5m/s2) t

  -20m/s / -5m/s2 = t

t = 4s

The time taken for the car to stop is 4s.

ii) distance =?

3rd equation of motion (2as = v2-u2)

2as=v2-u2

s= v2-u2/2a

  = 0 - (2om/s)2 /2(-5m/s2)

  = -400 m/s2 / 50 m/s2

  = -8 m

Thus, the car travelled a distance of 8 m in the opposite direction.

Hope this helps!