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?

Asked by julianabalaswamy762 | 6th Apr, 2020, 05:27: PM

Expert Answer:

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

Answered by Shiwani Sawant | 6th Apr, 2020, 06:08: PM