A car moving with a velocity \(6.25 \mathrm{~ms}^{-1}\) is decelerated with \(2.5 \sqrt{\mathrm{v}}…
A car moving with a velocity \(6.25 \mathrm{~ms}^{-1}\) is decelerated with \(2.5 \sqrt{\mathrm{v}} \mathrm{ms}^{-2}\) ( \(v\) is instantaneous velocity). Time taken by the car to come to rest is
\(2 \mathrm{~s}\)
\(3 \mathrm{~s}\)
\(2.5 \mathrm{~s}\)
\(4 \mathrm{~s}\)
Solution
Given, \(u=6.25 \mathrm{~ms}^{-1}\) and \(a=-2.5 \sqrt{v} \mathrm{~ms}^{-2}\), \(v=\) instantaneous velocity
As we know,
\(a=\frac{d v}{d t}\)
So,
\(\frac{d v}{d t}=-2.5 \sqrt{v}\)
Integrate on the both sides, we get
\(\begin{array}{llrl}
\Rightarrow & \quad \int_{6.25}^0 \frac{1}{\sqrt{v}} d v & =-2.5 \int d t \\
\Rightarrow & {[2 \sqrt{v}]_{6.25}^0} & =-2.5 t \\
\Rightarrow & t & =\frac{2 \times 2.5}{2.5}=2 \mathrm{~s}
\end{array}\)
Hence, the correct option is (a).