An object- moving with a speed of 6-25 m-s- is decelerated a rate given by dfrac - dv - dt - -2-5sqrt - - - where v is the instantaneous speed- The time taken by object to come rest would be-1 s2 s4 s8 s

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Dvdt-x2212-2-5-x221A-v-x222B-dv-x221A-v-x222B-x2212-2-5dt-2-x221A-v-06-25-x2212-2-5-t-t0-x2212-2-xD7-2-5-x2212-2-5tt-2sec

An object, moving with a speed of $6.25 m / s$ , is decelerated at a rate given by $\frac{d v}{d t} = - 2.5 \sqrt{v}$ , where $v$ is the instantaneous speed. The time taken by object to come rest would be:
$1 s$
$2 s$
$4 s$
$8 s$

$\frac{d v}{d t} = - 2.5 \sqrt{v} \int \frac{d v}{\sqrt{v}} = \int - 2.5 d t {[2 \sqrt{v}]}_{6.25}^{0} = - 2.5 {[t]}_{0}^{t} - 2 \times 2.5 = - 2.5 t t = 2 s e c$

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by dvdt=−2.5√v, where v is the instantaneous speed. The time taken by object to come rest would be:1 s2 s4 s8 s

dvdt=−2.5√v∫dv√v=∫−2.5dt[2√v]06.25=−2.5[t]t0−2×2.5=−2.5tt=2sec

An object, moving with a speed of $6.25 m / s$ , is decelerated at a rate given by $\frac{d v}{d t} = - 2.5 \sqrt{v}$ , where $v$ is the instantaneous speed. The time taken by object to come rest would be:
$1 s$
$2 s$
$4 s$
$8 s$

$\frac{d v}{d t} = - 2.5 \sqrt{v} \int \frac{d v}{\sqrt{v}} = \int - 2.5 d t {[2 \sqrt{v}]}_{6.25}^{0} = - 2.5 {[t]}_{0}^{t} - 2 \times 2.5 = - 2.5 t t = 2 s e c$