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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 | 06 Apr, 2020, 05:27: PM
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
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 | 06 Apr, 2020, 06:08: PM

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