A body travelling with uniform acceleration crosses two points A and B with velocities $20 \mathrm{~m} /…
A body travelling with uniform acceleration crosses two points A and B with velocities $20 \mathrm{~m} / \mathrm{s}$ and $30 \mathrm{~m} / \mathrm{s}$ respectively. The speed of the body at mid point of A and B is (nearly)
$25 \mathrm{~m} / \mathrm{s}$
$25.5 \mathrm{~m} / \mathrm{s}$
$24 \mathrm{~m} / \mathrm{s}$
$10 \sqrt{6} \mathrm{~m} / \mathrm{s}$
Solution
Let the acceleration of the body be ' $a$ ' and the distance between points A and B be ' d '.
$\begin{aligned}
& \therefore \quad \mathrm{v}^2-\mathrm{u}^2=2 \mathrm{ad} \quad \ldots\left(\text { from } \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as}\right) \\
& \mathrm{ad}=\frac{\mathrm{v}^2-\mathrm{u}^2}{2}=\frac{(30)^2-(20)^2}{2} \\
&=250
\end{aligned}$ When the body is at the mid-point of $A B$, then speed of the body is
$\begin{aligned}
& \mathrm{v}_1^2-(20)^2=2 \mathrm{a}\left(\frac{\mathrm{~d}}{2}\right) \\
& \mathrm{v}_1^2-400=\mathrm{ad} \\
& \mathrm{v}_1=\sqrt{650} \\
& \mathrm{v}_1=25.5 \mathrm{~m} / \mathrm{s}
\end{aligned}$
.