91Ó°ÊÓ

$$\begin{gathered} {\mathrm{v}}_0=24.6\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{f}}=0\mathrm{m}/\mathrm{s} \\ \mathrm{a}=-4.92\mathrm{m}/\mathrm{s}\mathrm{Here}\mathrm{negative}\mathrm{sign}\mathrm{because}\mathrm{of}\mathrm{deceleration}. \end{gathered}$$

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Kinematic equations can be used to find the time taken by the car to stop. Another kinematic equation would give the distance traveled by the car in that time. Further, the displacement and velocity for different time intervalscan be found to plot the graph.

Formula:

The final velocity in kinematic equations can be expressed as,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{a}×\mathrm{t}\left(1\right) \\ {{\mathrm{v}}_{\mathrm{f}}}^2={{\mathrm{v}}_0}^2+2\mathrm{a}×\mathrm{x} \\ \left(2\right) \\ \mathrm{Where}{\mathrm{v}}_0\mathrm{is}\mathrm{the}\mathrm{initial}\mathrm{velocity}\mathrm{a}\mathrm{is}\mathrm{an}\mathrm{acceleration}\mathrm{t}\mathrm{is}\mathrm{the} \\ \mathrm{time}\mathrm{and}\mathrm{x}\mathrm{is}\mathrm{the}\mathrm{displacement}. \end{gathered}$$

$$\begin{gathered} \mathrm{Using}\mathrm{equation}\left(\mathrm{i}\right) \\ {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{a}×\mathrm{t} \\ 0=24.6+(-4.92)×\mathrm{t} \\ \mathrm{t}=5.0\mathrm{s} \end{gathered}$$

$$\begin{gathered} \mathrm{Using}\mathrm{eq}+\mathrm{uation}\left(\mathrm{ii}\right), \\ {{\mathrm{v}}_{\mathrm{f}}}^2={{\mathrm{v}}_0}^2+2\mathrm{a}×\mathrm{x} \\ \mathrm{x}=\frac{{{\mathrm{v}}_{\mathrm{f}}}^2-{{\mathrm{v}}_0}^2}{2×\mathrm{a}} \\ =\frac{0^2-24.6^2}{2×(-4.92)} \\ =61.5\mathrm{m} \end{gathered}$$