Q. A body covers $40m$ in first $5 \, sec$ and then covers $65 \, m$ in next $5 \, sec$ . Find the initial velocity of the body if it is moving with uniform acceleration.
NTA AbhyasNTA Abhyas 2020
Solution:
We know, $S_{1}=ut+\frac{1}{2}at^{2}$ ... $\left(\right.i\left.\right)$
$t_{1}=t_{2}=t$ (given)
and $v=u+at$
Now $S_{2}=vt+\frac{1}{2}at^{2}$
putting value of $v$ in above equation $S_{2}=\left(u + a t\right)t+\frac{1}{2}at^{2}=ut+at^{2}+\frac{1}{2}at^{2}$ ... $\left(\right.ii\left.\right)$
Equation $\left(ii\right)-\left(i\right)=S_{2}-S_{1}=at^{2}$
$\Rightarrow a=\frac{S_{2} - S_{1}}{t^{2}}=\frac{65 - 40}{\left(5\right)^{2}}=1m/s^{2}$
From equation $\left(\right.i\left.\right)$ , we get.
$S_{1}=ut+\frac{1}{2}at^{2}$
$\Rightarrow 40=5u+\frac{1}{2}\times 1\times 25$
$\Rightarrow 5u=27.5$
$\therefore u=5.5m/s$
We know, $S_{1}=ut+\frac{1}{2}at^{2}$ ... $\left(\right.i\left.\right)$
$t_{1}=t_{2}=t$ (given)
and $v=u+at$
Now $S_{2}=vt+\frac{1}{2}at^{2}$
putting value of $v$ in above equation $S_{2}=\left(u + a t\right)t+\frac{1}{2}at^{2}=ut+at^{2}+\frac{1}{2}at^{2}$ ... $\left(\right.ii\left.\right)$
Equation $\left(ii\right)-\left(i\right)=S_{2}-S_{1}=at^{2}$
$\Rightarrow a=\frac{S_{2} - S_{1}}{t^{2}}=\frac{65 - 40}{\left(5\right)^{2}}=1m/s^{2}$
From equation $\left(\right.i\left.\right)$ , we get.
$S_{1}=ut+\frac{1}{2}at^{2}$
$\Rightarrow 40=5u+\frac{1}{2}\times 1\times 25$
$\Rightarrow 5u=27.5$
$\therefore u=5.5m/s$