A cyclist starts from rest and moves with a constant acceleration of 1 m/s².   A boy who is 48  m behind the cyclist starts moving with a  constant velocity of 10 m/s.  After how much time the boy meets the cyclist? 

            $$\begin{gathered} Let the boy meet the cyclist at B. \\ Boy has uniform motion : ut = total dis\tan ce \\ 10 t = 48 + d ---\left(1\right) \\ Cyclist has acceleration motion :\to initial velocity u=0;acceleration a =1m/s^2;dis\tan ce s=d;substitute in, \\ s=ut+\frac{1}{2}a{t}^2 \\ d=\frac{1}{2}\times 1\times t^2---\left(2\right) \end{gathered}$$$$\begin{gathered} \mathrm{substitute} \mathrm{d} \mathrm{value} \mathrm{from} \mathrm{equation} \left(1\right) \\ 10\mathrm{t}-48=\frac{1}{2}{\mathrm{t}}^2 \\ \Rightarrow {\mathrm{t}}^2-20\mathrm{t}+96=0 \\ (\mathrm{t}-8)( \mathrm{t}-12)=0 \\ \mathrm{The} \mathrm{boy} \mathrm{meets} \mathrm{the} \mathrm{cyclist} \mathrm{at} \mathrm{t} = 8 \sec , 12 \sec \end{gathered}$$

$$\begin{gathered} \mathrm{At} \mathrm{t} = 8 \sec , \mathrm{velocity} \mathrm{of} \mathrm{cyclist} =1 \times 8=8\mathrm{m}/\mathrm{s} \\ \mathrm{velocity} \mathrm{of} \mathrm{boy} >\mathrm{velocity} \mathrm{of} \mathrm{cyclist}, \mathrm{boy} \mathrm{overtakes} \mathrm{cyclist} \end{gathered}$$

$$\begin{gathered} \mathrm{At} \mathrm{t} = 12 \sec , \mathrm{velocity} \mathrm{of} \mathrm{cyclist} =1\times 12=12\mathrm{m}/\mathrm{s} \\ \mathrm{velocity} \mathrm{of} \mathrm{cyclist} >\mathrm{velocity} \mathrm{of} \mathrm{boy}, \mathrm{cyclist} \mathrm{overtakes} \mathrm{boy} \end{gathered}$$