A vehicle moving with velocity 2m/s can be stopped over a distance of 2m. Keeping the retarding force constant if kinetic energy is doubled what is the distance covered before it comes to rest?

Asked by Tejas Kulkarni | 23rd Dec, 2013, 01:10: PM

Expert Answer:

Given that,

u1 = 2 m/s

s = 2 m

Now v1 = 0

v12 – u12 = 2as

=> 0 – 22 = 2×a×2

=> a = -1 ms-2

The Kinetic energy of the body,

K1= ½ m(2)2

    = 2m

Now if Kinetic energy is doubled then,

K2 = 4m

Let velocity of the body become u2

½ mu2 2 = 4m

=> u2 = 2√2

Let s’ = distance covered before it comes to rest

v22 – u22 = 2as'

=> 02 – (2√2)2

= 2×(-1)×s’

=> s' = 4 m

Answered by Komal Parmar | 23rd Dec, 2013, 02:34: PM

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