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The kinetic energy of a car is doubled when its velocity is increased by $1 \mathrm{~ms}^{-1}$. Then the initial velocity of the car is
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Verified Answer
The correct answer is:
$(1+\sqrt{2}) \mathrm{ms}^{-1}$
Let initial velocity $=\mathrm{v}$
When velocity is increases by $1 \mathrm{~m} / \mathrm{s}$
$\mathrm{v}^{\prime}=\mathrm{v}+1$
then $\mathrm{KE}=2 \mathrm{E}$
$\frac{2 \mathrm{E}}{\mathrm{E}}=\frac{\frac{1}{2} \mathrm{~m}(\mathrm{v}+1)^2}{\frac{1}{2} \mathrm{mv}^2}$
$2 \mathrm{v}^2=(\mathrm{v}+1)^2$
$2 v^2=v^2+1+2 v$
$\begin{aligned}
& v^2-2 v-1=0 \Rightarrow v=\frac{2 \pm \sqrt{4+4}}{2} \\
& =\frac{2 \pm 2 \sqrt{2}}{2}=1+\sqrt{2} \mathrm{~m} / \mathrm{s}
\end{aligned}$
When velocity is increases by $1 \mathrm{~m} / \mathrm{s}$
$\mathrm{v}^{\prime}=\mathrm{v}+1$
then $\mathrm{KE}=2 \mathrm{E}$
$\frac{2 \mathrm{E}}{\mathrm{E}}=\frac{\frac{1}{2} \mathrm{~m}(\mathrm{v}+1)^2}{\frac{1}{2} \mathrm{mv}^2}$
$2 \mathrm{v}^2=(\mathrm{v}+1)^2$
$2 v^2=v^2+1+2 v$
$\begin{aligned}
& v^2-2 v-1=0 \Rightarrow v=\frac{2 \pm \sqrt{4+4}}{2} \\
& =\frac{2 \pm 2 \sqrt{2}}{2}=1+\sqrt{2} \mathrm{~m} / \mathrm{s}
\end{aligned}$
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