Question

A body starting from rest was observed to travel 20 m in 1 sec and 40 m during the next. How far had it travelled before the first observation was taken.

Answer 1

$$\begin{gathered} \mathrm{Case}\mathrm{I}:(\mathrm{t}=0\mathrm{to}\mathrm{t}) \\ \mathrm{Initial}\mathrm{velocity}=0 \\ \mathrm{Final}\mathrm{velocity}=\mathrm{v} \\ \mathrm{Distance}=\mathrm{x} \\ {\mathrm{v}}^2-0=2\mathrm{ax} \\ \mathrm{x}=\frac{{\mathrm{v}}^2}{2\mathrm{a}} \\ \mathrm{Case}\mathrm{II}:(\mathrm{t}\mathrm{to}\mathrm{t}+1) \\ \mathrm{Initial}\mathrm{velocity}=\mathrm{v} \\ \mathrm{Final}\mathrm{velocity}={\mathrm{v}}_1 \\ \mathrm{Distance}=20\mathrm{m} \\ \mathrm{Time}=1\mathrm{s} \\ 20=\mathrm{v}\left(1\right)+\frac{1}{2}\mathrm{a}{\left(1\right)}^2 \\ 20=\mathrm{v}+0.5\mathrm{a}.......\left(1\right) \\ {\mathrm{v}}_1=\mathrm{v}+\mathrm{a}\left(1\right) \\ {\mathrm{v}}_1=\mathrm{v}+\mathrm{a} \\ \mathrm{Case}\mathrm{III}:(\mathrm{t}+1\mathrm{to}\mathrm{t}+2) \\ \mathrm{Initial}\mathrm{velocity}={\mathrm{v}}_1 \\ \mathrm{Distance}=40\mathrm{m} \\ \mathrm{Time}=1\mathrm{s} \\ 40={\mathrm{v}}_1\left(1\right)+\frac{1}{2}\mathrm{a}{\left(1\right)}^2 \\ 40=\mathrm{v}+\mathrm{a}+0.5\mathrm{a} \\ 40=\mathrm{v}+1.5\mathrm{a}.......\left(2\right) \\ \left(2\right)-\left(1\right): \\ 20=\mathrm{a} \\ \mathrm{a}=20\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{v}=20-0.5\mathrm{a} \\ =20-0.5\times 20 \\ =10\mathrm{m}/\mathrm{s} \\ \mathrm{Hence}, \\ \mathrm{x}=\frac{{\mathrm{v}}^2}{2\mathrm{a}} \\ =\frac{{10}^2}{2\times 20} \\ =2.5\mathrm{m} \end{gathered}$$