A car velocity increases uniformly from 6m/s to 20m/s while covering 70m find the acceleration and the time taken.

Time taken = `"Distance"/"average velocity"`

= `\frac{70 m}{13 \frac{m}{s}}` = 5.38 s

Acceleration = (14 m/s)/5.38 s

= 2.6 m/s^2

Thus, the acceleration and stopping distance = 2.6 m/s^2