Question

A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.

Answer 1

When the car accelerates uniformly, the speed increases linearly: $v(t) = v_0 + a t$, from where given $v_0 = 18.5 \frac{m}{s}$ and $v(2.47) = 46.1\frac{m}{s}$ the acceleration is $a = \frac{v(2.47) - v_0}{2.47} \approx 11.17\frac{m}{s^2}$.

The law of motion with constant acceleration is $s(t) = v_0 t + \frac{a t^2}{2}$, and substituting $a = \frac{v(t) - v_0}{t}$, obtain $s(t) = v_0 t + \frac{v(t) - v_0}{2} t \approx 79.8 m$.