Question

An aeroplane on a runway, starts from rest and picks up a velocity of 180 km/h and takes off. In doing so, it covers a runway of 1.5 km. Calculate the time in which it takes off?

Answer 1

Step-1: Given data:

Initial velocity(u) = 0

Final velocity (v)= $180\raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$hr$}\right.=180\times \frac{1000m}{3600s}=50\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Distance= 1.5 km

Step-2

As we know that,

$$\begin{gathered} v^2=u^2+2\times a\times s \\ {\left(50\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2=0+2\times a\times 1500m \\ a=0.83\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \\ v=u+a\times t \\ 50\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.=0+0.83\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\times t \\ t=60s \end{gathered}$$

Hence, 60 s is answer.