a car moving with a velocity of 20m/s stopped in a distance of 40 m . if the same car is travelling at double the velocity the distance travelled by it for same retardation is?

Initial velocity of car, u = 20 m/s 

The final velocity will be, v= 0 as the car comes to rest after travelling distance of 40 m (which will be after application of brakes)

 

By third equation of motion we know, 

v² = u² + 2 as 

0 = (20)² + 2 x a x (40)

- 80 a = 400

Thus, a = - 5 m/s² 

Negative sign indicates that it is a retardation. 

When same car is moving with double the velocity 

Initial velocity, u = 2 x 20  =40 m/s 

Thus, 

0 = (40)² x 2 x (-5) x s 

10 s = 1600

s = 160 m 

Hence, if the same car is travelling at double the velocity the distance travelled by it for same retardation is 160 m