Question

A bus starts from rest with a constant acceleration of 5m/s². At the same time a car traveling with a constant velocity of 50m/s overtakes and passes the bus. (i) Find at what distance will the bus overtakes the car? (ii) How fast will the bus be travelling then?

Answer 1

$$\begin{gathered} \mathrm{Let} \\ \mathrm{t}=\mathrm{Time}\mathrm{to}\mathrm{overtake} \\ \left(\mathrm{i}\right) \\ \mathrm{Condition}\mathrm{to}\mathrm{overtake}: \\ \mathrm{Distance}\mathrm{travelled}\mathrm{by}\mathrm{car}=\mathrm{Distance}\mathrm{travelled}\mathrm{by}\mathrm{bus} \\ {\left[\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\right]}_{\mathrm{car}}={\left[\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\right]}_{\mathrm{bus}} \\ {\left[\left(0\right)\mathrm{t}+\frac{1}{2}\left(5\right){\mathrm{t}}^2\right]}_{\mathrm{car}}={\left[\left(50\right)\mathrm{t}+\frac{1}{2}\left(0\right){\mathrm{t}}^2\right]}_{\mathrm{bus}} \\ 2.5{\mathrm{t}}^2=50\mathrm{t} \\ 2.5\mathrm{t}=50 \\ \mathrm{t}=20\mathrm{s} \\ \mathrm{Hence}, \\ \mathrm{distance}=\mathrm{ut} \\ =50\times 20 \\ \mathbf{=}\mathbf{1000}\mathbf{}\mathbf{m} \\ \left(\mathrm{ii}\right) \\ \mathrm{Velocity}\mathrm{of}\mathrm{bus}: \\ \mathrm{v}=\mathrm{at} \\ =5\times 20 \\ \mathbf{=}\mathbf{100}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s} \end{gathered}$$