Chapter 11, Problem 78PQ

February 3, 2009, was a very snowy day along Interstate 69 just outside of Indianapolis, Indiana. As a result of the slippery conditions and low visibility (50 yards or less), there was an enormous accident involving approximately 30 vehicles, including cars, tractor-trailers, and even a fire truck. Many witnesses said that people were driving too fast for the conditions and were too close together. In this problem, we explore two rules of thumb for driving in such conditions. The first is to drive at a speed that is half of what it would be in ideal conditions. The other is the “8-second” rule: Watch the vehicle in front of you as it passes some object such as a street sign, and you should pass that same object 8 seconds later. On a dry road, the 8-second rule is replaced by a 3-second rule. a. Assume vehicles on a slippery interstate highway follow both rules. What is the distance between the vehicles? b. If a driver followed the first rule of thumb, driving at a lower speed, but used the 3-second rule instead of the 8-second rule, what is the distance between the vehicles? How does that distance compare with the visibility on the day of the accident? c. Suppose drivers do not follow either rule of thumb for slippery conditions. What is the distance between vehicles? How does that distance compare with the visibility on that day? d. Suppose a driver was not obeying either rule of thumb when she sees a tractor-trailer that stopped on the highway. She presses on her brakes, locking the wheels, and her car crashes into the truck. Estimate the magnitude of the impulse exerted on her car. e. Estimate the impulse on the car in part (d) had the driver followed both rules of thumb for slippery conditions instead of ignoring them.

To determine

The distance between the vehicles when running through the slippery road.

Answer to Problem 78PQ

The distance between the vehicles when running through the slippery road is $\underset{\_}{120\text{m}}$.

Explanation of Solution

Assume that the speed of the vehicles in highway is $65\text{mph}$ ($29\text{m/s}$), For a slippery road, the speed of the vehicle would be $15\text{m/s}$. Given that the one vehicle is $8\text{s}$ behind the other. The spacing between the vehicles when one is $8\text{s}$ behind the other, will be equal to the distance travelled by the vehicle in that time duration.

Write the expression for the distance travelled by the vehicle (the distance between the vehicles).

  $\Delta x=v\Delta t$                                                                                                                      (I)

Here, $\Delta x$ is the distance, $v$ is the speed of the vehicle, and $\Delta t$ is the time duration.

Conclusion:

Substitute $15\text{m/s}$ for $v$, and $8\text{s}$ for $\Delta t$ in equation (I) to find $\Delta x$.

  $\begin{array}{c}\Delta x=\left(15\text{m/s}\right)\left(8\text{s}\right)\\ =120\text{m}\end{array}$

Therefore, the distance between the vehicles when running through the slippery road is $\underset{\_}{120\text{m}}$.

To determine

The distance between the vehicles according to the $3\text{s}$ rule, and its comparison with the visibility on the day of the accident.

Answer to Problem 78PQ

The distance between the vehicles according to the $3\text{s}$ rule is $\underset{\_}{45\text{m}}$. This distance is about equal to the visibility on the day of the accident.

Explanation of Solution

Given that the visibility on the day of accident is $50\text{yards}$ or less.

Equation (I) gives the distance between the vehicles.

  $\Delta x=v\Delta t$

When the $3\text{s}$ rule is applied, the time duration is $3\text{s}$. That is one vehicle is $3\text{s}$ behind the other.

Conclusion:

Substitute $15\text{m/s}$ for $v$, and $3\text{s}$ for $\Delta t$ in equation (I) to find $\Delta x$.

  $\begin{array}{c}\Delta x=\left(15\text{m/s}\right)\left(3\text{s}\right)\\ =45\text{m}\end{array}$

The visibility on the day of accident in the units of meters is,

  $\begin{array}{c}50\text{yards}=50\text{yards}\times \frac{\text{0}\text{.9144}\text{m}}{1\text{yard}}\\ \approx 45\text{m}\end{array}$

This indicates that the distance between the vehicles when the $3\text{s}$ rule is applied, is equal to the visibility on the day of the accident.

Therefore, the distance between the vehicles according to the $3\text{s}$ rule is $\underset{\_}{45\text{m}}$. This distance is about equal to the visibility on the day of the accident.

To determine

The distance between the vehicles according to the $3\text{s}$ rule, and its comparison with the visibility on the day of the accident when the vehicle travels at $30\text{m/s}$.

Answer to Problem 78PQ

The distance between the vehicles according to the $3\text{s}$ rule, and its comparison with the visibility on the day of the accident when the vehicle travels at $30\text{m/s}$ is $\underset{\_}{90\text{m}}$. This is twice the visibility distance on the day of accident.

Explanation of Solution

Given that the visibility on the day of accident is $50\text{yards}$ ($45\text{m}$) or less.

Equation (I) gives the distance between the vehicles.

  $\Delta x=v\Delta t$

In this case the vehicle is moving with speed of $30\text{m/s}$ and follows the $3\text{s}$ rule.

Conclusion:

Substitute $30\text{m/s}$ for $v$, and $3\text{s}$ for $\Delta t$ in equation (I) to find $\Delta x$.

  $\begin{array}{c}\Delta x=\left(30\text{m/s}\right)\left(3\text{s}\right)\\ =90\text{m}\end{array}$

This indicates that the distance between the vehicles when moving with speed of $30\text{m/s}$ and following $3\text{s}$ rule, is twice the visibility on the day of the accident. The driver would not be able to see the car moving front of his.

Therefore, the distance between the vehicles according to the $3\text{s}$ rule, and its comparison with the visibility on the day of the accident when the vehicle travels at $30\text{m/s}$ is $\underset{\_}{90\text{m}}$. This is twice the visibility distance on the day of accident.

To determine

The magnitude of impulse on the car.

Answer to Problem 78PQ

The magnitude of impulse on the car is $\underset{\_}{5\times {10}^4\text{kg}\cdot \text{m/s}}$.

Explanation of Solution

According to the energy conservation, the kinetic energy of the vehicle at the crash equals the initial energy minus the energy dissipated by friction after applying the brakes.

  $K_f=K_i-\Delta E_{\text{th}}$                                                                                                            (II)

Here, $K_f$ is the final kinetic energy of the car, $K_i$ is the initial kinetic energy of the car, and $\Delta E_{\text{th}}$ is the change in thermal energy.

Write the expression for the kinetic energy of the car at the moment of crash.

  $K_f=\frac{1}{2}m{v}_c^2$                                                                                                                (III)

Here, $m$ is the mass of the car, and $v_c$ is the speed of the car at the moment of crash.

Write the expression for the initial kinetic energy of the car.

  $K_i=\frac{1}{2}m{v}_i^2$                                                                                                                (IV)

Here, $v_i$ is the initial speed of the car.

Write the expression for the change in thermal energy.

  $\Delta E_{\text{th}}={\mu }_{k}mg\Delta x$                                                                                                         (V)

Here, ${\mu }_k$ is the coefficient of kinetic friction, $g$ is the acceleration due to gravity, and $\Delta x$ is the distance moved.

Use equation (III), (IV), and (V) in (II) and solve for $v_c$.

  $\begin{array}{c}\frac{1}{2}m{v}_c^2=\frac{1}{2}m{v}_i^2-{\mu }_{k}mg\Delta x\\ v_c=\sqrt{v_i^2-2{\mu }_{k}g\Delta x}\end{array}$                                                                                         (VI)

The impulse on the car is equal to the change in momentum. Since the truck stops the car after collision, the final momentum is zero. Thus, the magnitude of impulse on the car will be equal to the momentum of the car at the moment of collision.

  $\left|I\right|=m{v}_c$                                                                                                                    (VII)

Here, $I$ is the impulse.

The mass of the car be $2000\text{kg}$, the initial speed be $30\text{m/s}$, the distance moved is $50\text{m}$, and the coefficient of kinetic friction between the ice and rubber is $0.15$.

Conclusion:

Substitute $30\text{m/s}$ for $v_i$, $0.15$ for ${\mu }_k$, $9.81{\text{m/s}}^2$ for $g$, and $50\text{m}$ for $\Delta x$ in equation (VI) to find $v_c$.

  $\begin{array}{c}{v}_c=\sqrt{{\left(30\text{m/s}\right)}^2-2\left(0.15\right)\left(9.81{\text{m/s}}^2\right)\left(50\text{m}\right)}\\ =27\text{m/s}\end{array}$

Substitute $27\text{m/s}$ for $v_c$ and $2000\text{kg}$ for $m$ in equation (VII) to find $\left|I\right|$.

  $\begin{array}{c}\left|I\right|=\left(2000\text{kg}\right)\left(27\text{m/s}\right)\\ =5.4\times {10}^4\text{kg}\cdot \text{m/s}\\ \approx 5\times {10}^4\text{kg}\cdot \text{m/s}\end{array}$

Therefore, the magnitude of impulse on the car is $\underset{\_}{5\times {10}^4\text{kg}\cdot \text{m/s}}$.

To determine

The magnitude of impulse on the car with rules for the slippery road is applied.

Answer to Problem 78PQ

The magnitude of impulse on the car with rules for the slippery road is applied is $\underset{\_}{2\times {10}^4\text{kg}\cdot \text{m/s}}$.

Explanation of Solution

When the rules for the slippery road is applied, the initial speed of the car would be $15\text{m/s}$.

Equation (VI) gives the speed of the car at the moment of crash.

  $v_c=\sqrt{v_i^2-2{\mu }_{k}g\Delta x}$

Equation (VII) gives the magnitude of impulse on the car.

  $\left|I\right|=m{v}_c$

Conclusion:

Substitute $15\text{m/s}$ for $v_i$, $0.15$ for ${\mu }_k$, $9.81{\text{m/s}}^2$ for $g$, and $50\text{m}$ for $\Delta x$ in equation (VI) to find $v_c$.

  $\begin{array}{c}{v}_c=\sqrt{{\left(15\text{m/s}\right)}^2-2\left(0.15\right)\left(9.81{\text{m/s}}^2\right)\left(50\text{m}\right)}\\ =9\text{m/s}\end{array}$

Substitute $9\text{m/s}$ for $v_c$ and $2000\text{kg}$ for $m$ in equation (VII) to find $\left|I\right|$.

  $\begin{array}{c}\left|I\right|=\left(2000\text{kg}\right)\left(9\text{m/s}\right)\\ =1.8\times {10}^4\text{kg}\cdot \text{m/s}\\ \approx 2\times {10}^4\text{kg}\cdot \text{m/s}\end{array}$

Therefore, the magnitude of impulse on the car with rules for the slippery road is applied is $\underset{\_}{2\times {10}^4\text{kg}\cdot \text{m/s}}$.