91Ó°ÊÓ

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

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

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

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