To stop a car, first you require a certain reaction time to begin braking; then the car slows at a constant rate. Suppose that the total distance moved by your car during these two phases is $\mathbf{56}\mathbf{.}\mathbf{7}$mwhen its initial speed is$\mathbf{80}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}$, and$\mathbf{24}\mathbf{.}\mathbf{4}$mwhen its initial speed is$\mathbf{48}\mathbf{.}\mathbf{3}\mathbf{km}\mathbf{/}\mathbf{h}$. What are (a) your reaction time and (b) the magnitude of the acceleration?

The distance moved by a car while braking with the initial speed${\mathrm{v}}_{01}=80.5\mathrm{km}/\mathrm{h}$in the first phase is ${\mathrm{x}}_1=56.7\mathrm{m}.$

The distance moved by a car while braking with the initial speed${\mathrm{v}}_{02}=48.3\mathrm{km}/\mathrm{h}$in the second phase is ${\mathrm{x}}_2=24.4\mathrm{m}$.

Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence motion. The equations which are used in the study are known as kinematic equations of motion. Using the kinematic equation, you can analyze the motion of a body.

The distance traveled by car in reaction time t_r, can be given as,

$∆\mathrm{x}={\mathrm{v}}_0{\mathrm{t}}_{\mathrm{r}}+{\mathrm{v}}_0{\mathrm{t}}_{\mathrm{b}}+\frac{1}{2}{\mathrm{at}}_{\mathrm{b}}^2$

After the breaks are applied, the car will start to decelerate, after time t_b, v= 0,

$$\begin{gathered} \mathrm{v}={\mathrm{v}}_0+{\mathrm{at}}_{\mathrm{b}} \\ 0={\mathrm{v}}_0+{\mathrm{at}}_{\mathrm{b}} \\ -{\mathrm{v}}_0={\mathrm{at}}_{\mathrm{b}} \\ {\mathrm{t}}_{\mathrm{b}}=-\frac{{\mathrm{v}}_0}{\mathrm{a}} \end{gathered}$$

Substituting the value of t_bin the equation of x

$$\begin{gathered} \mathrm{x}={\mathrm{v}}_0{\mathrm{t}}_{\mathrm{r}}-\frac{{\mathrm{v}}_0^2}{\mathrm{a}}+\frac{1}{2}\times \mathrm{a}\times \frac{{\mathrm{v}}_0^2}{{\mathrm{a}}^2} \\ ={\mathrm{v}}_0{\mathrm{t}}_{\mathrm{r}}-\frac{{\mathrm{v}}_0^2}{\mathrm{a}}+\frac{1}{2}\frac{{\mathrm{v}}_0^2}{\mathrm{a}} \\ ={\mathrm{v}}_0{\mathrm{t}}_{\mathrm{r}}-\frac{1}{2}\frac{{\mathrm{v}}_0^2}{\mathrm{a}} \\ \mathrm{Now},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}, \\ {\mathrm{x}}_1=56.7\mathrm{m}, \\ {\mathrm{x}}_2=24.4\mathrm{m} \\ \mathrm{Also},\mathrm{convert}\mathrm{the}\mathrm{velocities}\mathrm{in}\mathrm{m}/\mathrm{s}. \\ {\mathrm{v}}_{01}=80.5\frac{\mathrm{km}}{\mathrm{hr}}\times \frac{1\mathrm{hr}}{3600\sec }\times \frac{1000}{1\mathrm{km}} \\ =13.4\mathrm{m}/\mathrm{s} \\ \mathrm{With}\mathrm{this}\mathrm{you}\mathrm{can}\mathrm{say}\mathrm{that}, \\ {\mathrm{x}}_1={\mathrm{v}}_0{\mathrm{t}}_{\mathrm{r}}-\frac{{\mathrm{v}}_{02}^2}{2\mathrm{a}} \\ \mathrm{Solving}\mathrm{the}\mathrm{above}\mathrm{two}\mathrm{equations}\mathrm{simultaneously}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{value}{\mathrm{t}}_{\mathrm{r}}\mathrm{of}\mathrm{a}, \\ {\mathrm{t}}_{\mathrm{r}}=\frac{{\mathrm{v}}_{02}^2{\mathrm{x}}_1-{\mathrm{v}}_{01}^2{\mathrm{x}}_2}{{\mathrm{v}}_{01}{\mathrm{v}}_{02}\left({\mathrm{v}}_{02}-{\mathrm{v}}_{01}\right)} \\ \left(\mathrm{i}\right) \\ \mathrm{a}=-\frac{1}{2}\times \left(\frac{{\mathrm{v}}_{02}{\mathrm{v}}_{01}^2-{\mathrm{v}}_{01}{\mathrm{v}}_{02}^2}{{\mathrm{v}}_2{\mathrm{x}}_1-{\mathrm{v}}_{01}{\mathrm{x}}_2}\right) \\ \left(\mathrm{ii}\right) \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{to}\mathrm{calculate}{\mathrm{t}}_{\mathrm{r}}. \\ {\mathrm{t}}_{\mathrm{r}}=\frac{\left(13.4^2\times 56.7\right)-\left(22.4^2\times 24.4\right)}{\left(22.4\times 13.4\right)\times \left(13.4-22.4\right)} \\ =0.74\mathrm{s} \\ \mathrm{So},\mathrm{the}\mathrm{reaction}\mathrm{time}\mathrm{is}0.74\sec . \end{gathered}$$

Now, substitute the values in equation (ii) to calculate .

$$\begin{gathered} \mathrm{a}=\left(-\frac{1}{2}\times \left(\frac{\left(13.4\times 22.4^2\right)-\left(22.4\times 13.4^2\right)}{\left(13.4\times 56.7\right)-\left(22.4\times 24.4\right)}\right)\right) \\ =-6.2\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{acceleration}\mathrm{is}-6.2\mathrm{m}/{\mathrm{s}}^2\mathrm{and}\mathrm{the}\mathrm{magnitude}\mathrm{is}6.2\mathrm{m}/{\mathrm{s}}^2. \end{gathered}$$