Question

A car moving under uniform acceleration covers a distance of $180m$ from point P to point Q in $6s$. The speed of the car as it passes through point $Q$ is $45m{s}^{-1}$. Calculate the speed and acceleration of the car at the time when it passes through point P.

Answer 1

Given that,

Speed of the Car, $V=45m/s$
Distance, $S=180m$
Time, $T=6s$

Initial Speed, $u=?$
Acceleration, $a=?$
We know that
$V=u+at$
$45=u+6a$ ....... (1)
By second equation of motion
$S=ut+\frac{1}{2}a{t}^2$
$180=6u+18a$ ...... (2)
Solving equation (1)and (2) we get
$u=15m/s$
$a=5m/s^2$