Q114P - Fundamentals Of Physics

Question

A car can be braked to a stop from the autobahn-like speed of ${}^{\mathbf{200}\mathbf{ }\mathbf{km}\mathbf{/}\mathbf{h}}$ in ${}^{\mathbf{170}\mathbf{ }\mathbf{m}}$. Assuming the acceleration is constant, find its magnitude in (a) SI units and (b) in terms of g. (c) How much time ${}^{{\mathbf{T}}_{\mathbf{b}}}$ is required for the braking? Your reaction time ${}^{{\mathbf{T}}_{\mathbf{r}}}$ is the time you require to perceive an emergency, move your foot to the brake, and begin the braking. If ${}^{{\mathbf{T}}_{\mathbf{r}}\mathbf{ }\mathbf{=}400 \mathrm{ms}}$, then (d) what is ${}^{{\mathbf{T}}_{\mathbf{b}}}$ in terms of ${}^{{\mathbf{T}}_{\mathbf{r}}}$ , and (e) is most of the full time required to stop spent in reacting or braking? Dark sunglasses delay the visual signals sent from the eyes to the visual cortex in the brain, increasing ${}^{{\mathbf{T}}_{\mathbf{r}}}$ . (f) In the extreme case in which ${}^{{\mathbf{T}}_{\mathbf{r}}}$ is increased by , how much farther does the car travel during your reaction time?

Step-by-Step Solution

Answer

1. The magnitude of the acceleration of the car is ${}^{\mathbf{9}\mathbf{.}\mathbf{08}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}}$.

2. The acceleration of the car is ${}^{\mathbf{0}\mathbf{.}\mathbf{926}\mathbf{g}}$.

3. The time required for braking the car is ${}^{\mathbf{6}\mathbf{.}\mathbf{12}\mathbf{ }\mathbf{s}}$. 

4. The time required for braking is ${}^{\mathbf{15}\mathbf{.}\mathbf{3}{\mathbf{T}}_{\mathbf{r}}}$.

5. Most of the time required to stop the car was used for braking.

6. For ${}^{\mathbf{500}\mathbf{ }\mathbf{ms}}$ of reaction time, the car will travel ${}^{\mathit{27}\mathit{.}\mathit{78}\mathit{ }\mathit{m}}$ farther.

Step 1: To understand the concept

The initial speed of the car, ${}^{{\mathbf{v}}_{\mathbf{0}}\mathbf{ }\mathbf{=}\mathbf{200}\mathbf{ }\mathbf{km}\mathbf{/}\mathbf{h}}$ 

The distance covered, ${}^{\mathbf{x}\mathbf{ }\mathbf{=}\mathbf{170}\mathbf{ }\mathbf{m}}$ 

Step 2: Understanding the kinematic equations

Kinematic equations describe the motion of an object with constant acceleration. These equations can be used to determine the acceleration, velocity or distance.

The expression for the kinematic equations of motion are given as follows: 

${}^{\mathit{v}\mathit{=}{\mathit{v}}_{\mathit{0}}\mathit{+}\mathit{a}\mathit{t}}$                                                                                            … (i)

${}^{{\mathit{v}}^{\mathbf{2}}\mathit{=}{\mathit{v}}_{\mathbf{0}}^{\mathbf{2}}\mathit{+}\mathit{2}\mathit{a}\mathit{x}}$                                                                                       … (ii)

Here, ${}^{{\mathbf{v}}_{\mathbf{o}}}$ is the initial velocity, ${}^{\mathbf{v}}$ is the final velocity, ${}^{\mathbf{t}}$ is the time, ${}^{\mathbf{a}}$ is the acceleration and ${}^{\mathbf{x}}$ is the displacement.

Step 3: (a) Determination of the magnitude of acceleration in SI units

Convert the speed of the car from km/h to m/s as: 

$$\begin{gathered} {\mathbf{v}}_{\mathbf{0}}\mathbf{ }\mathbf{=}\mathbf{200}\mathbf{ }\mathbf{km}\mathbf{/}\mathbf{h} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{200}\mathbf{\times }\frac{\mathbf{5}}{\mathbf{18}}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{55}\mathbf{.}\mathbf{56}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s} \end{gathered}$$ 

Using equation (ii), the acceleration is calculated as follows: 

$$\begin{gathered} \mathbf{a}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{-}{\mathbf{v}}_{\mathbf{0}}^{\mathbf{2}}}{\mathbf{2}\mathbf{ }\mathbf{x}} \\ \mathbf{ }\mathbf{ }\mathbf{=}\frac{{\left(0 m/s\right)}^{\mathbf{2}}\mathbf{ }\mathbf{-}{\left(55.56 m/s\right)}^{\mathbf{2}}}{\mathbf{2}\mathbf{\times }\mathbf{170}\mathbf{ }\mathbf{m}} \\ \mathbf{ }\mathbf{ }\mathbf{=}\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}} \end{gathered}$$ 

Since it is a negative value, it can also be called as deceleration.

Thus, the magnitude of acceleration is ${}^{\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}}$. 

Step 4: (b) Determination of the magnitude of acceleration in terms of g

As ${}^{\mathit{g}\mathbf{=}\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}}$, the magnitude of acceleration of the car is,

$$\begin{gathered} \mathbf{a}\mathbf{=}\mathbf{\left(}\frac{\mathbf{9}\mathbf{.}\mathbf{08}}{\mathbf{9}\mathbf{.}\mathbf{8}}\mathbf{\right)}\mathbf{g} \\ \mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{926}\mathbf{g} \end{gathered}$$

Thus, the acceleration of the car is ${}^{\mathbf{0}\mathbf{.}\mathbf{926}\mathbf{g}}$. 

Step 5: (c) Determination of the time required for braking

Using equation (i), time required for braking is calculated as follows: 

$$\begin{gathered} \mathbf{T}\mathbf{=}\frac{\mathbf{V}\mathbf{-}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{a}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{55}\mathbf{.}\mathbf{56}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{08}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{6}\mathbf{.}\mathbf{12}\mathbf{ }\mathbf{s} \end{gathered}$$ 

Thus, the time required for braking the car is ${}^{\mathbf{ }\mathbf{6}\mathbf{.}\mathbf{12}\mathbf{ }\mathbf{s}}$

Step 6: (d) Determination of T b in terms of T r

The reaction time is, 

$$\begin{gathered} {\mathbf{T}}_{\mathbf{r}}\mathbf{ }\mathbf{=}\mathbf{400}\mathbf{ }\mathbf{ms} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{4} \end{gathered}$$

The braking time in terms of reaction time is, 

$$\begin{gathered} \frac{{\mathbf{T}}_{\mathbf{b}}}{{\mathbf{T}}_{\mathbf{r}}}\mathbf{=}\frac{\mathbf{6}\mathbf{.}\mathbf{12}\mathbf{ }\mathbf{s}}{\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{ }\mathbf{s}} \\ {\mathbf{T}}_{\mathbf{b}}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{3}{\mathbf{T}}_{\mathbf{t}} \end{gathered}$$

Thus, the time required for braking the car is ${}^{\mathbf{15}\mathbf{.}\mathbf{3}{\mathbf{T}}_{\mathbf{t}}}$.

Step 7: (e) Determination where most of the time used for

Since, the braking time is much higher than the reaction time $\left({}^{T_b=15.3T_r}\right)$ therefore, the braking required most of the time for the car to stop. 

Step 8: (f) Determination how much farther does the car travel during reaction time

The reaction time ${}^{{\mathbf{T}}_{\mathbf{r}}}$ is increased by ${}^{\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{s}}$.

So,

$$\begin{gathered} {\mathbf{T}}_{\mathbf{r}}\mathbf{=}\mathbf{400}\mathbf{ }\mathbf{ms}\mathbf{+}\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{100}\mathbf{ }\mathbf{ms} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{s} \end{gathered}$$

The distance traveled by the car in this period would be,

Thus, the car will travel further by ${}^{\mathbf{27}\mathbf{.}\mathbf{78}\mathbf{ }\mathbf{m}}$ in the reaction time.