A BULLET STOPS AFTER TRAVELLING A DISTANCE OF XM UNDER THE ACTION OF A UNIFORM RETARDING FORCE .HOW FAR WILL IT GO BEFORE BEING STOPPED BY THE SAME FORCE IF ITS VELOCITY IS DOUBLED?

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Solution

Let the initial velocity be ‘u’
It travels a distance ‘x’ before coming to rest.
Using,
v² = u² + 2ax
=> 0 = u² + 2ax
=> a = -u²/(2x)
If the initial velocity is ‘2u’ then,
v² = (2u)² + 2a(x^/)
=> 0 = 4u² – [2u²/(2x)](x^/)
=> x^/ = 4x
Thus, when the initial velocity is double the distance traveled before coming to rest is ‘4x’.

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