In 3.0 seconds, the braking of an automobile reduces its speed from 59 miles per hour to 20 miles per hour. How far does...

$$\begin{gathered} \mathrm{Solution}:-\mathrm{Given} \mathrm{that} \\ \mathrm{time} \left(\mathrm{t}\right)=3 \mathrm{s} \\ \mathrm{Initial} \mathrm{speed} \left(\mathrm{u}\right)=59 \mathrm{miles} \mathrm{per} \mathrm{hour} \\ \mathrm{Final} \mathrm{speed} \left(\mathrm{V}\right)=20 \mathrm{miles} \mathrm{per} \mathrm{hour} \\ 1 \mathrm{mile}=1609 \mathrm{m} \end{gathered}$$

$$\begin{gathered} \mathrm{Initial} \mathrm{speed}\left(\mathrm{u}\right)=\frac{59\times 1609}{3600}\mathrm{m}/\mathrm{s}=26.37 \mathrm{m}/\mathrm{s} \\ \mathrm{Final} \mathrm{speed} \left(\mathrm{V}\right)=\frac{20\times 1609}{3600} \mathrm{m}/\mathrm{s}=8.94 \mathrm{m}/\mathrm{s} \\ \mathrm{Now}, \mathrm{using} \mathrm{equation} \mathrm{of} \mathrm{motion}; \\ \mathrm{V}=\mathrm{u}+\mathrm{at} \\ 8.94=26.37+\mathrm{a}\times 3 \\ \mathrm{a}=-5.81 \mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Again} \mathrm{using} \mathrm{equation} \mathrm{of} \mathrm{motion}; \\ \mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{S}=26.37\times 3+\frac{1}{2}\times (-5.81)\times 3^2 \\ \mathrm{S}=52.965 \mathrm{m} \\ \mathbf{Hence}\mathbf{,}\mathbf{ }\mathbf{distance}\mathbf{ }\mathbf{travelled}\mathbf{ }\mathbf{by}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{vehicle}\mathbf{ }\mathbf{during}\mathbf{ }\mathbf{3}\mathbf{ }\mathbf{s}\mathbf{=}\mathbf{52}\mathbf{.}\mathbf{965}\mathbf{ }\mathbf{m} \end{gathered}$$