PhysicsMedium65×since 2002Q7883An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by : dvdt=−2.5v{{dv} \over {dt}} = - 2.5\sqrt vdtdv=−2.5v where v is the instantaneous speed. The time taken by the object, to come to rest, would be :A2 sB4 sC8 sD1 sCheck answerSkip