Q75P - Fundamentals Of Physics

Question

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 $\mathbf{56}\mathbf{.}\mathbf{7}$m when its initial speed is $\mathbf{80}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{km}\mathbf{/}\mathbf{h}$, and $\mathbf{24}\mathbf{.}\mathbf{4}$m when 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?

Step-by-Step Solution

Answer

(a) Your reaction time is 0.74 s

(b) The magnitude of the acceleration is $6.2 \mathrm{m}/{\mathrm{s}}^2$.

Step 1: Given data

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 ${\mathrm{x}}_2=24.4 \mathrm{m}$.

Step 2: Kinematic Equations of Motion

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.

Step 3: (a) Calculations for the reaction time

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_b in 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}$$

Step 4: (b) Calculations for acceleration

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

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