a body moving with uniform retardation covers 3km before the speed is reduced to half of its initial value. It comes to rest in another distance of how many metres?

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let initial velocity = v

final velocity = v/2

distance travelled = 3 m

${V}^{2}-{u}^{2}=2as\phantom{\rule{0ex}{0ex}}\frac{{v}^{2}}{4}-{v}^{2}=2a\times 3\phantom{\rule{0ex}{0ex}}-\frac{3{v}^{2}}{4}=6a\phantom{\rule{0ex}{0ex}}a=\frac{-{v}^{2}}{8}km/{s}^{2}$

Now in the second case

initial velocity = v/2

final velocity = 0

let distance travelled = x

$weget\phantom{\rule{0ex}{0ex}}-\frac{{v}^{2}}{4}=2ax\phantom{\rule{0ex}{0ex}}onsubstitutingthevalueofaweget\phantom{\rule{0ex}{0ex}}x=1km$

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