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Chapter 7: Problem 57

Legal Physics. In an auto accident, a car hit a pedestrian and the driver then slammed on the brakes to stop the car. During the subsequent trial, the driver's lawyer claimed that he was obeying the posted 35 -mph speed limit, but the legal speed was too high to allow him to see and react to the pedestrian in time, You have been called in as the state's expert witness. Your investigation of the accident found that the skid marks made while the brakes were applied were 280 \(\mathrm{ft}\) long, and the tread on the tires produced a coefficient of kinetic friction of 0.30 with the road. (a) In your testimony in court, will you say that the driver was obeying the posted speed? You must be able to back up your conclusion with clear reasoning because one of the lawyers will surely cross-examine you. (b) If the driver's speeding ticket were \(\$ 10\) for each mile per hour he was driving above the posted speed limit, would he have to pay a tine? If so, how much would it be?

Short Answer

Expert verified
The driver was speeding, traveling at approximately 50.1 mph. The fine for speeding is $151.

Step by step solution

01

Convert Speed Limit to ft/s

First, let's convert the speed limit from miles per hour to feet per second, since 1 mile is 5280 feet and 1 hour is 3600 seconds. Therefore, 35 mph can be converted using the formula:\[35 \text{ mph} = 35 \times \frac{5280}{3600} \approx 51.33 \text{ ft/s}\]
02

Calculate Deceleration Using Friction

The force of friction, which causes deceleration, is calculated using the formula that includes the coefficient of kinetic friction \( \mu_k \):\[ f_k = \mu_k \cdot m \cdot g \]where \( f_k \) is the kinetic friction force, \( m \) is the mass of the car, and \( g \) is the acceleration due to gravity (32.2 ft/s虏). The deceleration \( a \) caused by this friction force is given by:\[ a = \mu_k \cdot g = 0.3 \times 32.2 \approx 9.66 \text{ ft/s}^2 \]
03

Use the Work-Energy Principle to Find Initial Speed

The work-energy principle relates initial speed (\( v_i \)), final speed (zero at stop), and distance traveled under constant deceleration:\[ v_f^2 = v_i^2 + 2ad \]Solving for \( v_i \) (initial speed) since \( v_f \) (final speed) is 0:\[ 0 = v_i^2 + 2(-9.66 \text{ ft/s}^2)(280 \text{ ft}) \]\[ v_i = \sqrt{-2ad} = \sqrt{2 \times 9.66 \times 280} \approx 73.45 \text{ ft/s} \]
04

Determine if Speeding Occurred

Compare the initial speed of 73.45 ft/s to the speed limit of 51.33 ft/s. Since 73.45 ft/s is greater than 51.33 ft/s, the driver was indeed speeding.
05

Calculate Speeding Fine

Convert 73.45 ft/s back to mph:\[ 73.45 \text{ ft/s} = 73.45 \times \frac{3600}{5280} \approx 50.1 \text{ mph} \]The driver was traveling 50.1 mph, which is 15.1 mph over the speed limit of 35 mph. Thus, the fine is:\[ \text{Fine} = 15.1 \times 10 = \$151 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coefficient of Kinetic Friction
The coefficient of kinetic friction is an important concept in physics, especially when analyzing situations involving motion and friction. It is a dimensionless value representing the ratio of the force of friction between two bodies and the force pressing them together.
The kinetic coefficient specifically applies when the bodies are sliding past each other, as opposed to the static coefficient of friction, which applies when the objects are not moving. In the context of the legal case described, the coefficient of kinetic friction was given as 0.30, indicating the amount of frictional resistance provided by the road surface against the tires of the car.
  • The coefficient of kinetic friction (\( \mu_k \)) is used to calculate the force of friction (\( f_k \)).
  • The formula is \( f_k = \mu_k \cdot m \cdot g \), where \( m \) is the mass of the car and \( g \) is the acceleration due to gravity.
  • This value is crucial for determining how quickly a vehicle can stop by overcoming kinetic friction.
Understanding this concept helps in evaluating how the car's speed and the road conditions interacted at the moment the brakes were applied.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics that connects the work done by forces to the change in kinetic energy of an object. This principle helps determine how the speed of an object is affected by the energy involved.
In the scenario presented, this principle was used to determine the car's initial speed before the brakes were applied.
  • The relationship is expressed with the equation: \( v_f^2 = v_i^2 + 2ad \), where \( v_f \) is the final speed (zero, in this case), \( v_i \) is the initial speed, \( a \) is acceleration (negative when decelerating), and \( d \) is the displacement (the length of the skid marks).
  • By rearranging the formula to solve for the initial speed, we can determine how fast the vehicle was moving before the frictional forces halted its motion.
The calculated initial speed confirmed whether the driver adhered to the speed limit.
Speed Conversion
Speed conversion is often necessary when working with different units of measurement, especially in kinematics problems. In this particular legal case, it was essential to convert the speed limit from miles per hour to feet per second.
  • This conversion is done because many physical calculations, such as those involving friction and acceleration, are set in standard units like feet per second squared for consistency.
  • For conversion, the formula \( \text{Miles per hour} \times \frac{5280 \, \text{feet}}{3600 \, \text{seconds}} \) is used, as 1 mile equals 5280 feet and 1 hour equals 3600 seconds. Hence, \( 35 \text{ mph} \) becomes approximately \( 51.33 \text{ ft/s} \).
This calculation set the basis for comparing the car's speed to the legal limit in the same unit of feet per second, ensuring accurate analysis.
Deceleration Calculation
Deceleration is a concept involving the rate at which an object slows down or decreases its speed. It is generally a negative acceleration. In this scenario, the deceleration was caused by the force of kinetic friction acting between the car's tires and the road.
  • The calculation of deceleration involves the coefficient of kinetic friction and the acceleration due to gravity. The formula is \( a = \mu_k \cdot g \), which, in this case, becomes \( 0.3 \cdot 32.2 \text{ ft/s}^2 = 9.66 \text{ ft/s}^2 \)
  • This value represents the rate at which the car slowed down once the brakes were applied.
  • Knowing the deceleration is essential to apply the work-energy principle to determine the car's initial speed.
Understanding deceleration helps in comprehending how the frictional forces influenced the outcome of the accident.

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Most popular questions from this chapter

A \(0.150-\mathrm{kg}\) block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is 1.20 \(\mathrm{m}\) above the floor. The spring has force constant 1900 \(\mathrm{N} / \mathrm{m}\) and is initially compressed 0.045 \(\mathrm{m} .\) The mass of the spring is negligible. The spring is released, and the block slides along the table, goes off the edge, and travels to the floor. If there is negligible friction between the block of ice and the tabletop, what is the speed of the block of ice when it reaches the floor?

CALC (a) Is the force \(\vec{F}=C y^{2} \hat{\jmath},\) where \(C\) is a negative constant with units of \(\mathrm{N} / \mathrm{m}^{2}\) , conservative or nonconservative? Justify your answer. (b) Is the force \(\vec{\boldsymbol{F}}=C y^{2 \hat{\boldsymbol{l}}}\) , where \(C\) is a negative constant with units of \(\mathrm{N} / \mathrm{m}^{2},\) conservative or nonconservative? Justify your answer.

Two blocks with different masses are attached to either end of a light rope that passes over a light, frictionless pulley suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended \(1.20 \mathrm{m},\) its speed is 3.00 \(\mathrm{m} / \mathrm{s} .\) If the total mass of the two blocks is 15.0 \(\mathrm{kg},\) what is the mass of each block?

A wooden rod of negligible mass and length 80.0 \(\mathrm{cm}\) is pivoted about a horizontal axis through its center. A white rat with mass 0.500 kg clings to one end of the stick, and a mouse with mass 0.200 kg clings to the other end. The system is released from rest with the rod horizontal. If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position?

CALC An object has several forces acting on it. One force is \(\vec{\boldsymbol{F}}=\alpha x y \hat{\boldsymbol{r}},\) a force in the \(x\) -direction whose magnitude depends on the position of the object. (Sce Problem \(6.98 . )\) The constant is \(\alpha=2.00 \mathrm{N} / \mathrm{m}^{2}\) . The object moves along the following path: \((1)\) It starts at the origin and moves along the \(y\) -axis to the point \(x=0\) , \(y=1.50 \mathrm{m} ;(2)\) it moves parallel to the \(x\) -axis to the point \(x=1.50 \mathrm{m}, y=1.50 \mathrm{m} ;(3)\) it moves parallel to the \(y\) -axis to the point \(x=1.50 \mathrm{m}, y=0 ;(4)\) it moves parallel to the \(x\) -axis back to the origin. (a) Sketch this path in the \(x y\) -plane. (b) Calculate the work done on the object by \(\vec{F}\) for each leg of the path and for the complete round trip. (c) Is \(\vec{F}\) conservative or nonconservative? Explain.
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