Kinematics

JEE Main level, single correct, medium

135 of 308

Question 135 of 308

An object, moving with a speed of 6.25 m/s6.25 \mathrm{~m} / \mathrm{s}, is decelerated at a rate given by :
dvdt=−2.5v\frac{\mathrm{dv}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}}
where vv is the instantaneous speed. The time taken by the object, to come to rest, would be:

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Answer: (A) 2 s2 \mathrm{~s}

dvdt=−2.5v\frac{d v}{d t}=-2.5 \sqrt{v}
Integrating the above equation.
⇒2v=−2.5t+C\Rightarrow 2 \sqrt{v}=-2.5 \mathrm{t}+\mathrm{C}
at t=0,v=6.25⇒C=5\mathrm{t}=0, \mathrm{v}=6.25 \Rightarrow \mathrm{C}=5
at v=0⇒t=52.5=2 sv=0 \Rightarrow t=\frac{5}{2.5}=2 \mathrm{~s}

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