A car slows down from 108 km/hr to 18 km/hr over a distance of 40 meter. If the brakes are applied with the same force . Calculate 

1) total time in which car comes to rest .

2) total distance travelled by it.

When, u = 108 km/h = 30 m/s

and v = 18 km/h = 5 m/s

s = 40 m 

 

We can calculate the acceleration using 3rd equation of motion: - 

 

v² - u² = 2as 

(5)² - (30)² = 2 a (40) 

 

25 - 900 = 80a

a = - 10.93 m/s² 

Negative sign indicates that it is retardation. 

 

i)Total time in which car comes to rest: - 

As the brakes are applied with same force to bring car to rest, acceleration remains same i.e. 10.93 m/s². 

 

v = 0, u = 30 m/s 

a = -10.93 m/s² 

 

Thus, 

v = u + at 

0 = 30 + (-10.93) t 

-30 = -10.93 t 

Thus, 

t =  2.74 s

 

ii) Distance travelled by car during this time is given by, 

 

v² - u² = 2as 

(0)² - (30)² = 2(-10.93) s

 

- 900 = 21.86 s 

s =  41.17 m