An object, moving with a speed of $6.25 \mathrm{~m} / \mathrm{s}$, is decelerated at a rate given by : $…

An object, moving with a speed of $6.25 \mathrm{~m} / \mathrm{s}$, is decelerated at a rate given by : $ \frac{\mathrm{dv}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}} $ where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be:
  1. $2 \mathrm{~s}$
  2. $4 \mathrm{~s}$
  3. $8 \mathrm{~s}$
  4. $1 \mathrm{~s}$

Solution

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

Asked in: JEE Main 2011

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