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

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Solution

Given that,

Mass = m

u₁ = 2 m/s

s = 2 m

and v = 0
Using equation,

v² – u² = 2as

=> 0 – 2² = 2×a×2

=> a = -1 ms^-2

KE of the body = ½ m(2)²

= 2m

Now KE if doubled then KE = 4m

Let velocity of the body become u₂

½ mu₂ ² = 4m

=> u₂ = 2√2

Let s’ be the distance at which it will stop

v² – u² = 2as’

=> 0² – (2√2)² = 2×(-1)×s’

=> s = 4 m

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