Question

A car moving with constant acceleration covered the distance between two points 60.0 m apart in 6.00 s. Its speed as it passes the second point was 15.0 m/s. What is the acceleration?

Answer 1

$v=15m/s,s=60m,t=6sv=u+at\Rightarrow vt=ut+a{t}^2....\left(1\right)s=ut+\frac{1}{2}a{t}^2.....\left(2\right)vt-s=\frac{1}{2}a{t}^{2}a=\frac{vt-s}{\frac{1}{2}{t}^2}=\frac{90-60}{18}=\frac{5}{3}m/s^2$