If initial velocity of an object is doubled, the stopping distance becomes 4 times. Could any one of you explain me this in detail using the equation S = u^2 / 2a?
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the initial velocity = u
the velocity gets double= 2u
the stopping distance becoming = 4 times of d
S = u^2 /2a
4d= 4u^2/2a
2a = 4u^2/4d
2a = u^2 / d
hope this will help you
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