A point moving with constant acceleration from $A$ to $B$ in the straight line $A B$ has velocities $u$ and $v$ at $A$ and $B$ respectively. Find its velocity at $C$ , the mid point of $A B$ .

\sqrt{\frac{v^{2} - u^{2}}{2}}

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\sqrt{\frac{v^{2} + u^{2}}{2}}

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\sqrt{\frac{v^{2} + u^{2}}{4}}

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\sqrt{\frac{v^{2} - u^{2}}{4}}

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Solution

The correct option is B $\sqrt{\frac{v^{2} + u^{2}}{2}}$
Let the distance between $A$ and $B$ = $s$ and as mentioned $C$ is the midpoint of $A$ and $B$ , thus distance $A C$ and $C B$ = $\frac{s}{2}$
From equation of motion we get distance $A B$ as
$v^{2} = u^{2} + 2 a s$
$s = \frac{v^{2} - u^{2}}{2 a}$ - ---------(1)
Let speed of the vehicle be $p m/s$ at point $C$
Again, from equation of motion we get,
$p^{2} = u^{2} + 2 a \frac{s}{2}$
as distance between $A$ and $C$ is $\frac{s}{2}$
$s = \frac{p^{2} - u^{2}}{a}$ -------(2)
Equating 1 and 2, we get
$\frac{p^{2} - u^{2}}{a} = \frac{v^{2} - u^{2}}{2 a}$
$p^{2} = \frac{v^{2} + u^{2}}{2} \Rightarrow p = \sqrt{\frac{v^{2} + u^{2}}{2}}$

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