A body moving with constant retardation 'a' losses 2/ 3 of its initial velocity V0 what is the distance covered by the body in the time during which this occurs?

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Solution

Dear student,

$U \sin g e q u a t i o n o f m o t i o n v^{2} = u^{2} - 2 a s l o o s e s 2 / 3 o f i n i t i a l v e l o c i t y s o f i n a l v e l o c i t y = 1 / 3 V_{0} (1 / 3 V_{0})^{2} = {V_{0}}^{2} - 2 a s 2 a s = {V_{0}}^{2} - \frac{{V_{0}}^{2}}{9} 2 a s = \frac{8 {V_{0}}^{2}}{9} s = \frac{4 {V_{0}}^{2}}{9 a}$

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A body moving with constant retardation 'a' losses 2/ 3 of its initial velocity V0 what is the distance covered by the body in the time during which this occurs?

Dear student, Using equation of motionv2 =u2-2aslooses 2/3 of initial velocity so final velocity = 1/3V0(1/3V0)2 =V02-2as2as =V02-V0292as=8V029s=4V029a Regards