In 4.0 seconds, the braking of an automobile reduces its speed from 58 miles per hour to 10 miles per hour. How far does...

$$\begin{gathered} \mathrm{Solution}:-\mathrm{Given} \mathrm{that} \\ \mathrm{time} \left(\mathrm{t}\right)=4 \mathrm{s} \\ \mathrm{Initial} \mathrm{speed} \left(\mathrm{u}\right)=58 \mathrm{miles} \mathrm{per} \mathrm{hour} \\ \mathrm{Final} \mathrm{speed} \left(\mathrm{V}\right)=10 \mathrm{miles} \mathrm{per} \mathrm{hour} \\ 1 \mathrm{mile}=1609 \mathrm{m} \end{gathered}$$

$$\begin{gathered} \mathrm{Initial} \mathrm{speed}\left(\mathrm{u}\right)=\frac{58\times 1609}{3600}\mathrm{m}/\mathrm{s}=25.923 \mathrm{m}/\mathrm{s} \\ \mathrm{Final} \mathrm{speed} \left(\mathrm{V}\right)=\frac{10\times 1609}{3600} \mathrm{m}/\mathrm{s}=4.47 \mathrm{m}/\mathrm{s} \\ \mathrm{Now}, \mathrm{using} \mathrm{equation} \mathrm{of} \mathrm{motion}; \\ \mathrm{V}=\mathrm{u}+\mathrm{at} \\ 4.47=25.923+\mathrm{a}\times 4 \\ \mathrm{a}=-5.36325 \mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Again} \mathrm{using} \mathrm{equation} \mathrm{of} \mathrm{motion}; \\ \mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{S}=25.923\times 4+\frac{1}{2}\times (-5.36325)\times 4^2 \\ \mathrm{S}=60.786 \mathrm{m} \\ \mathbf{Hence}\mathbf{,}\mathbf{ }\mathbf{distance}\mathbf{ }\mathbf{travelled}\mathbf{ }\mathbf{by}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{vehicle}\mathbf{ }\mathbf{during}\mathbf{ }\mathbf{4}\mathbf{ }\mathbf{s}\mathbf{=}\mathbf{60}\mathbf{.}\mathbf{786}\mathbf{ }\mathbf{m} \end{gathered}$$