An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by: dvdt=-2.5vwhere v is the instantaneous speed. The time taken (in seconds) by the object to come to rest would be

dvdt=-2.5v or dvv=-2.5dtIntegrating,∫6.250 dvv=-2.5∫0t dtor 2v1/26.250=-2.5t⇒t=-2×(6.25)1/2-2.5=2 s

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An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by:
$\frac{dv}{dt} = - 2 .5 \sqrt{v}$
where $v$ is the instantaneous speed. The time taken (in seconds) by the object to come to rest would be

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$\frac{dv}{dt} = - 2 .5 \sqrt{v} or \frac{dv}{\sqrt{v}} = - 2 .5 dt$

Integrating,

$\int_{6 .25}^{0} \frac{dv}{\sqrt{v}} = - 2 .5 \int_{0}^{t} dt$

$or 2 {[v^{1 / 2}]}_{6 .25}^{0} = - 2 .5 t$

$\Rightarrow t = \frac{- 2 \times (6 .25)^{1 / 2}}{- 2 .5} = 2 s$

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