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A man is at a distance of $6 \mathrm{~m}$ from a bus. The bus begins to move with a constant acceleration of $3 \mathrm{~m} \mathrm{~s}^{-2}$. In order to catch the bus, the minimum speed with which the man should run towards the bus is
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The correct answer is:
$6 \mathrm{~m} \mathrm{~s}^{-1}$
If the man did not run, the bus would be at a distance $s_1$ at time $t$ given by
$s_1=6+\frac{1}{2} a t^2=6+\frac{1}{2} \times 3 \times t^2=6+\frac{3}{2} t^2$
If $v$ is the speed of man, he would cover a distance $s_2=v t$ in time $t$. To catch the bus, $s_1=s_2$
$6+\frac{3}{2} t^2=v t \quad$ or $\quad t^2-\frac{2 v}{3} t+4=0$
which gives $t=\frac{2 v}{6} \pm \frac{1}{2}\left[\frac{4 v^2}{9}-16\right]^{1 / 2}$
Now, $t$ will be real if $\left(\frac{4 v^2}{9}-16\right)$ is positive or zero. Minimum $v$ corresponds to $\frac{4 v^2}{9}-16=0$ which gives $v=6 \mathrm{~m} \mathrm{~s}^{-1}$.
$s_1=6+\frac{1}{2} a t^2=6+\frac{1}{2} \times 3 \times t^2=6+\frac{3}{2} t^2$
If $v$ is the speed of man, he would cover a distance $s_2=v t$ in time $t$. To catch the bus, $s_1=s_2$
$6+\frac{3}{2} t^2=v t \quad$ or $\quad t^2-\frac{2 v}{3} t+4=0$
which gives $t=\frac{2 v}{6} \pm \frac{1}{2}\left[\frac{4 v^2}{9}-16\right]^{1 / 2}$
Now, $t$ will be real if $\left(\frac{4 v^2}{9}-16\right)$ is positive or zero. Minimum $v$ corresponds to $\frac{4 v^2}{9}-16=0$ which gives $v=6 \mathrm{~m} \mathrm{~s}^{-1}$.
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