A car moves linearly with uniform retardation . If the car covers 40 m in the last 2 seconds of  its motion , what is the velocity of the car at the beginning of  the last second ?

Dear Student, 
Please find below the solution to the asked query: 
$$\begin{gathered} \mathrm{Distance} \mathrm{covered} \mathrm{in} \mathrm{last} \mathrm{two} \mathrm{seconds}: \\ {\mathrm{s}}_{\mathrm{n}}+{\mathrm{s}}_{\mathrm{n}-1}=40 \mathrm{meters} \\ \mathrm{u}+\mathrm{a}\left(\mathrm{n}-\frac{1}{2}\right)+\mathrm{u}+\mathrm{a}\left(\mathrm{n}-1-\frac{1}{2}\right)=40 \\ 2\mathrm{u}+\mathrm{a}\left(\mathrm{n}-\frac{1}{2}+\mathrm{n}-\frac{3}{2}\right)=40 \\ 2\mathrm{u}+\mathrm{a}(2\mathrm{n}-2)=40 \\ \mathrm{u}+\mathrm{a}\left(\mathrm{n}-1\right)=20 \\ \mathrm{Velocity} \mathrm{at} \mathrm{the} \mathrm{beginning} \mathrm{of} \mathrm{last} \mathrm{second}: \\ \mathrm{v}=\mathrm{u}+\mathrm{a}(\mathrm{n}-1) \\ \mathbf{=}\mathbf{20}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s} \end{gathered}$$
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