Question

An object moving with a speed of 6.25 m/s,6.25 m/s, is decelerated at a rate given by dvdt=−2.5v√,dvdt=−2.5v, where v is the instantaneous speed. The time taken by the object, to come to rest, would be

Answer 1

Initial velocity = $v_{0}$ = 6.25 m/s

$\frac{dv}{dt} = -2.5v$

$\frac{dv}{v} = -2.5dt$

Integrating, we get:

$[lnv]^v_{v_{0}} = -2.5(t - 0)$

So, $lnv - lnv_{0} = -2.5t$

$ln\frac{v}{v_{0}} = -2.5t$

$\frac{v}{v_{0}} = e^{-2.5t}$

$v = v_{0}e^{-2.5t}$

Object at rest => v = 0.

So, $v_{0}e^{-2.5t} = 0$

$e^{-2.5t} = 0$

For that, t must tend to infinity.

Hence, the object will never truly come to rest.