Question

A motor bike, initially moving at 18 km/h, accelerates at a rate of 5 m/s² for 5 sec. Calculate the distance covered by it from start.

Answer 1

Step-1:

Given data:

Initial velocity(u) = $18\raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$hr$}\right.=18\times \frac{1000m}{3600s}=5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Accelaration (a)= 5 $\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$

Time(t)=5 sec

Step-2

As we know that,

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ s=5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\times 5s+\frac{1}{2}\times 5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\times {(5s)}^2 \\ s=87.5m \end{gathered}$$

Hence, the distance covered by it from start is 87.5 m.