You are arguing over a cell phone while trailing an unmarked police car by $\mathbf{25}\mathbf{\hspace{0.33em}}\mathbf{m}$; both your car and the police car are traveling at $\mathbf{110}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}\mathbf{.}$.Your argument diverts your attention from the police car for $\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{s}$, (long enough for you to look at the phone and yell, “I won’t do that!”). At the beginning of that $\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{s}$,the police officer begins braking suddenly at $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$(a)What is the separation between the two cars when your attention finally returns? Suppose that you take another$\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{s}$ to realize your danger and begin braking. (b)If you too brake at$\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$,what is your speed when you hit the police car?

Initial Separation between the two cars$\mathrm{d}=25\mathrm{m}$

Deceleration of the police car $5\mathrm{m}/{\mathrm{s}}^2$

Reaction time 0.4 sec.

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration.

Using the formula for the first kinematic equation, the speed of the police car before the braking can be found. When the collision occurs at time t, displacement $\mathrm{x}\text{'}=\mathrm{x}"$, from which the speed of the car can be obtained.

Formula:

The general velocity distance relation is expressed by,

$$\begin{gathered} \mathrm{Distance}=\left(\mathrm{velocity}\right)\times \left(\mathrm{time}\right) \\ \mathrm{v}={\mathrm{v}}_0+\mathrm{at} \\ \mathrm{x}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Speed}=110\mathrm{km}/\mathrm{h} \\ =30.56\mathrm{m}/\mathrm{s} \end{gathered}$$

During your car travels a distance

$$\begin{gathered} \mathrm{Distance}\left({\mathrm{d}}_2\right)=\left(30.56\mathrm{m}/\mathrm{s}\right)\left(2\mathrm{s}\right)-\frac{1}{2}\left(5\mathrm{m}/{\mathrm{s}}^2\right){\left(2\mathrm{s}\right)}^2 \\ =51.12\mathrm{m} \end{gathered}$$

During this phase, the separation between your car and police car is reduced by,

$$\begin{gathered} {\mathrm{d}}_1-{\mathrm{d}}_2=61.12\mathrm{m}-51.12\mathrm{m} \\ ∆\mathrm{d}=10\mathrm{m} \end{gathered}$$

Initially, both the cars were separated by 25 m, Because of the situation above, the separation is reduced by 10 m. Therefore, the distance by which the cars are now separated is,

$$\begin{gathered} ∆\mathrm{D}=\mathrm{Initial}\mathrm{Separation}-\mathrm{Reduction}\mathrm{in}\mathrm{Separation} \\ =\mathrm{d}-∆\mathrm{d} \\ =25\mathrm{m}-10\mathrm{m} \\ =15\mathrm{m} \end{gathered}$$

Therefore, both the cars are separated by 15 m.

For b part, total time is

$$\begin{gathered} =2.0\sec +0.4\sec \left(\mathrm{reaction}\mathrm{time}\right) \\ =2.4\sec \end{gathered}$$

So distance traveled by your car is

$$\begin{gathered} \mathrm{x}=30.56\mathrm{m}/\mathrm{s}\times 2.4\mathrm{s} \\ =73.33\mathrm{m} \end{gathered}$$

And distance travelled by the police car is,

$$\begin{gathered} \mathrm{x}"=\left(30.56\mathrm{m}/\mathrm{s}\right)\times \left(2.4\mathrm{s}\right)-\frac{1}{2}\left(5\mathrm{m}/{\mathrm{s}}^2\right)\times {\left(2.4\mathrm{s}\right)}^2 \\ =58.93\mathrm{m} \end{gathered}$$

Initially, both the cars were separated by 25 m,now the separation is reduced by 14.4 m Thus, the gap between two cars at the start of braking is 10.6 m .The speed of police car at this point is

$$\begin{gathered} \mathrm{v}={\mathrm{v}}_0+\mathrm{at} \\ =30.56\mathrm{m}/\mathrm{s}-\left(5\mathrm{m}/{\mathrm{s}}^2\right)\times \left(2.4\mathrm{s}\right) \\ =18.56\mathrm{m}/\sec \end{gathered}$$

The collision occurs at time t’, when

$\mathrm{x}\text{'}=\mathrm{x}"$

We have,

$\mathrm{\Delta x}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$

Considering the position of your car at the time of braking as the origin, we have

$$\begin{gathered} \mathrm{x}"-0=\left(30.56\mathrm{m}/\mathrm{s}\right)\times \left(\mathrm{t}\text{'}-{\mathrm{t}}_0\right)\mathrm{s}-\frac{1}{2}\left(5\mathrm{m}/{\mathrm{s}}^2\right)\times {\left[\left(\mathrm{t}\text{'}-{\mathrm{t}}_0\right)\mathrm{s}\right]}^2 \\ \mathrm{x}"=\left(30.56\mathrm{m}/\mathrm{s}\right)\times \left(\mathrm{t}\text{'}-{\mathrm{t}}_0\right)\mathrm{s}-\frac{1}{2}\left(5\mathrm{m}/{\mathrm{s}}^2\right)\times {\left[\left(\mathrm{t}\text{'}-{\mathrm{t}}_0\right)\mathrm{s}\right]}^2 \end{gathered}$$

Since the police car is ahead of your car by ,we have

localid="1656999695367" $$\begin{gathered} \mathrm{x}"-10.6\mathrm{m}=\left(+18.56\mathrm{m}/\mathrm{s}\right)\times \left(\mathrm{t}\text{'}-{\mathrm{t}}_0\right)\mathrm{s}-\frac{1}{2}\left(5\mathrm{m}/{\mathrm{s}}^2\right)\times {\left[\left(\mathrm{t}\text{'}-{\mathrm{t}}_0\right)\mathrm{s}\right]}^2 \\ \mathrm{x}"=10.6\mathrm{m}+\left(18.56\mathrm{m}/\mathrm{s}\right)\times \left(\mathrm{t}\text{'}-{\mathrm{t}}_0\right)\mathrm{s}-\frac{1}{2}\left(5\mathrm{m}/{\mathrm{s}}^2\right)\times {\left[\left(\mathrm{t}\text{'}-{\mathrm{t}}_0\right)\mathrm{s}\right]}^2 \\ \mathrm{x}\text{'}=\mathrm{x}" \end{gathered}$$

Solving the above equation, we get

localid="1656999030601" $$\begin{gathered} 10.6\mathrm{m}=\left(30.56-18.56\right)\mathrm{m}/\mathrm{s}\times \left(\mathrm{t}\text{'}-{\mathrm{t}}_0\right)\mathrm{s} \\ \left(\mathrm{t}\text{'}-{\mathrm{t}}_0\right)=0.883\mathrm{s} \end{gathered}$$

Therefore, your speed at this time is,

localid="1656999587008" $$\begin{gathered} \mathrm{v}={\mathrm{v}}_0+\mathrm{a}\left(\mathrm{t}\text{'}-{\mathrm{t}}_0\right) \\ =30.56\mathrm{m}/\mathrm{s}-\left(5\mathrm{m}/{\mathrm{s}}^2\right)\times \left(0.88\mathrm{s}\right) \\ =26\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of your car when it hits the police car is $\mathrm{v}=26\mathrm{m}/\mathrm{s}\mathrm{Or}94\mathrm{km}/\mathrm{h}.$