Question

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

• A. $\sqrt{\frac{v^2-u^2}{2}}$
• B. $\sqrt{\frac{v^2+u^2}{2}}$
• C. $\sqrt{\frac{v^2+u^2}{4}}$
• D. $\sqrt{\frac{v^2-u^2}{4}}$

Answer 1

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