Question

A car moving with a velocity of 54 km/h accelerates uniformly at the rate of 2 ms^-2. Calculate the distance travelled from the place where acceleration began to that where the velocity reaches 73km/h. And the time taken to cover this distance.

Answer 1

$s=\frac{v^2-v_0^2}{2a}=\frac{20.28^2-15^2}{2\cdot 2}=46.57\ (m)$ . Answer

$v=v_0+at\to t=(v-v_0)/a=(20.28-15)/2=2.64\ (s)$ . Answer