A car moving with uniform acceleration covers the distance of $200 \mathrm{~m}$ in first 2 seconds and the…
A car moving with uniform acceleration covers the distance of $200 \mathrm{~m}$ in first 2 seconds and the distance of $220 \mathrm{~m}$ in next 4 seconds. The velocity of the car after 7 seconds is
$10 \mathrm{~ms}^{-1}$
$20 \mathrm{~ms}^{-1}$
$15 \mathrm{~ms}^{-1}$
$30 \mathrm{~ms}^{-1}$
Solution
Distance travelled in first 2 seconds $=200 \mathrm{~m}$
$S=u t+\frac{1}{2} a t^2$
$200=2 u+\frac{1}{2} \times a \times 2^2$
$u+a=100$ ...(1)
Distance travelled in next 4 seconds $=220 \mathrm{~m}$ Total distance travelled in first 6 second $=420 \mathrm{~m}$
$\begin{aligned} & 420=6 u+\frac{1}{2} \times 9 \times 6^2 \\ & 420=6 u+18 a\end{aligned}$
$\mathrm{u}+3 \mathrm{a}=70$ ...(2)
Solving equation (1) and (2), we have
$\begin{aligned} & \mathrm{a}=-15 \mathrm{~m} / \mathrm{s}^2, \mathrm{u}=115 \mathrm{~m} / \mathrm{s} \\ & \mathrm{v}=\mathrm{u}+\mathrm{at} \\ & =115-15 \times 7 \\ & =115-105=10 \mathrm{~m} / \mathrm{s}\end{aligned}$