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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 \(\mathrm{mi}\) h speed limit, but that 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 be was driving above the posted speed limit, would he have to pay a fine? If so, how much would it be?

Short Answer

Expert verified
(a) The driver was speeding at approximately 50.1 mi/h. (b) The fine would be $151.

Step by step solution

01

Convert Velocity Units

First, let's find the initial velocity of the car just before braking in feet per second, since the skid marks and friction are given in feet and seconds. To convert 35 miles per hour to feet per second, use the conversion:\[ 35 \, \text{mi/h} = 35 \, \text{mi/h} \times \frac{5280 \, \text{ft/mi}}{3600 \, \text{s/h}} \approx 51.33 \, \text{ft/s} \].
02

Determine Stopping Distance

The stopping distance formula involves the initial velocity \(v\), the acceleration \(a\), and the stopping distance \(d\) due to braking:\[ d = \frac{v^2}{2a} \].We need to find the actual velocity \(v\) using the given distance of 280 ft and the acceleration.
03

Calculate Acceleration due to Friction

The acceleration \(a\) is given by the formula \(a = \mu g\), where \(\mu = 0.30\) is the coefficient of kinetic friction and \(g = 32.2 \, \text{ft/s}^2\) is the acceleration due to gravity. Therefore:\[ a = 0.30 \times 32.2 = 9.66 \, \text{ft/s}^2 \].
04

Calculate Initial Velocity

With \(d = 280 \, \text{ft}\) and \(a = 9.66 \, \text{ft/s}^2\), rearrange the stopping distance equation to solve for \(v\):\[ v^2 = 2ad \].\[ v^2 = 2 \times 9.66 \times 280 \approx 5409.6 \],\[ v = \sqrt{5409.6} \approx 73.54 \, \text{ft/s} \].
05

Compare Speeds with Speed Limit

Convert the calculated velocity back to miles per hour:\[ v = 73.54 \, \text{ft/s} \times \frac{3600 \, \text{s/h}}{5280 \, \text{ft/mi}} \approx 50.1 \, \text{mi/h} \].Since 50.1 miles per hour is greater than 35 miles per hour, the driver was not obeying the speed limit.
06

Determine the Fine

Calculate how much over the speed limit the driver was going:\[ 50.1 \, \text{mi/h} - 35 \, \text{mi/h} = 15.1 \, \text{mi/h} \].The fine is \\(10 for each mile per hour over:\[ \text{Fine} = 15.1 \, \text{mi/h} \times 10 \, \text{\\)/mi/h} = \$151 \].

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Key Concepts

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

Kinetic Friction
When it comes to understanding what happened during the car's braking, kinetic friction plays a crucial role. Every time two surfaces slide against each other, friction is created. In this case, it was between the car's tires and the road. Kinetic friction is a type of friction that occurs when objects are moving relative to each other. It can sometimes slow things down significantly.

Kinetic friction is determined by a coefficient, noted as \(\mu\), which in this exercise was given as 0.30. - The coefficient depends on the characteristics of the surfaces in contact. - Frictional force is calculated using the formula: \( f_k = \mu F_n \), where \( F_n \) is the normal force.
The coefficient of kinetic friction tells us how "grippy" the surfaces are together. It's an essential component because during an emergency stop, the car’s ability to come to a halt is directly influenced by this friction force. Understanding kinetic friction helps answer the question: Was the driver really able to stop fast enough given his speed?
Motion
Motion is about an object changing its position over time. In physics, describing motion involves talking about speed, velocity, and acceleration. In this scenario, we're particularly interested in how the car moved just before and as it skidded to a halt on the road.

- The car's initial motion was driven by the driver's speed, which was the speed at which he was traveling before he applied the brakes. - The subsequent motion, or the car's deceleration, was heavily dictated by the application of kinetic friction against the tires.
When trying to understand whether the driver was speeding, it’s all about evaluating his motion through the stopping distance and determining whether he had enough time and distance to stop the vehicle safely. This evaluation might also take into consideration the laws of motion, such as Newton's laws, to describe and predict how the car moved and eventually stopped.
Velocity
Velocity is a vector quantity, which means it has both magnitude and direction. Often confused with speed, velocity tells us not just how fast an object is moving, but in which direction. In this case, we’re interested in the car’s velocity just before it began to skid.

- Initially, the velocity of the car just before braking needed to be calculated in consistent units (feet per second) to find out how fast the car was moving. - Using this value, further calculations show how velocity affected the car’s capability to stop within the road's confines.
By employing the relationship of the car's velocity and the coefficient of kinetic friction, we can calculate if the driver could have reasonably stopped the vehicle in time and within a safe distance. Calculating velocity accurately helps break down whether the car was indeed going beyond the legal limit.
Stopping Distance
Stopping distance is the total distance a vehicle travels before it comes to a full stop after the driver applies the brakes. This concept is crucial in understanding whether the car's actual speed was too high for safe stopping at the given conditions of friction and road surface.

- Stopping distance is determined by multiple factors, including the initial velocity of the vehicle and the friction between the tires and the road.- The formula used to calculate stopping distance: \( d = \frac{v^2}{2a} \), incorporates both the car's speed and its deceleration.
Forensic calculations using the stopping distance help ascertain how much time and space it took for the car to come to rest. When skid marks stretch for 280 ft, determining the initial speed using the friction coefficient can scientifically support whether or not the claimed speed limit was obeyed. The evaluated stopping distance provides crucial evidence to verify the truth of the driver’s claim.

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Most popular questions from this chapter

An object moving in the \(x y\) -plane is acted on by a conservative force described by the potential energy function \(U(x, y)=\) \(\alpha\left(1 / x^{2}+1 / y^{2}\right),\) where \(\alpha\) is a positive constant. Derive an expression for the force expressed in terms of the unit vectors \(\hat{i}\) and \(\hat{y}\) .

You and three friends stand at the corners of a square whose sides are 8.0 \(\mathrm{m}\) long in the middle of the gym floor, as shown in Fig. \(7.26 .\) You take your physics book and push it from one person to the other. The book has a mass of 1.5 \(\mathrm{kg}\) , and the coefficient of kinetic friction between the book and the floor is \(\mu_{\mathrm{x}}=0.25 .\) (a) The book slides from you to Beth and then from Beth to Carlos, along the lines connecting these people. What is the work done by friction during this displacement? (b) You slide the book from you to Carlos along the diagonal of the square. What is the work done by friction during this displacement? (c) You slide the book to Kim who then slides it back to you. What is the total work done by friction during this motion of the book? (d) Is the friction force on the book conservative or nonconservative? Explain.

A 1.20 \(\mathrm{kg}\) piece of cheese is placed on a vertical spring of negligible mass and force constant \(k=1800 \mathrm{N} / \mathrm{m}\) that is compressed 15.0 \(\mathrm{cm}\) . When the spring is released, how high does the cheese rise from this initial position? (The cheese and the spring are not attached.)

A \(75-\mathrm{kg}\) roofer climbs a vertical 7.0 -m ladder to the flat roof of a house. He then walks 12 \(\mathrm{m}\) on the roof, climbs down another vertical \(7.0-\mathrm{m}\) ladder, and finally walks on the ground back to his starting point. How much work is done on him by gravity (a) as he climbs up; \((b)\) as he climbs down; (c) as he walks on the roof and on the ground? (d) What is the total work done on him by gravity during this round trip? On the basis of your answer to part (d), would you say that gravity is a conservative or nonconservative force? Explain.

Up and Down the Hill. A \(28-\mathrm{kg}\) rock approaches the foot of a hill with a speed of 15 \(\mathrm{m} / \mathrm{s}\) . This hill slopes upward at a constant angle of \(40.0^{\circ}\) above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20 , respectively. (a) Use energy conservation to find the maximum height above the foot of the hill reached by the rock. (b) Will the rock remain at rest at its highest point, or will it slide back down the hill? (c) If the rock does slide back down, find its speed when it returns to the bottom of the hill.

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