Question

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 by the object, to come to rest, would be then

• A. $2s$
• B. $4s$
• C. $8s$
• D. $1s$

Answer 1

The correct option is A $2s$

$\frac{dv}{dt}=-2.5\sqrt{v}{v}^{-1/2}dv=-2.5dtIntegartingbothsideswithlimits6.25and0,asthebodyisreducesitsvelocity.{\int }_{6.25}^0{v}^{-1/2}dv={\int }_0^t-2.5dt(2v^{1/2}{)}_{6.25}^0=-2.5(t{)}_0^{t}2\{0-(\frac{5}{2}\left)\right\}=-2.5tt=2seconds$