/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 86 You are called as an expert witn... [FREE SOLUTION] | 91Ó°ÊÓ

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You are called as an expert witness in a trial for a traffic violation. The facts are these: A driver slammed on his brakes and came to a stop with constant acceleration. Measurements of his tires and the skid marks on the pavement indicate that he locked his car's wheels, the car traveled 192 ft before stopping, and the coefficient of kinetic friction between the road and his tires was 0.750. He was charged with speeding in a 45-mi/h zone but pleads innocent. What is your conclusion: guilty or innocent? How fast was he going when he hit his brakes?

Short Answer

Expert verified
Guilty; the driver was speeding at approximately 65.45 mi/h.

Step by step solution

01

Understand the Given Problem

We need to determine the initial speed of the car when the driver applied the brakes. We know that the car's wheels locked up and skidded 192 feet to a stop. The coefficient of kinetic friction is given as 0.750, and the allowed speed limit is 45 mi/h.
02

Convert Units

Convert the speed limit from miles per hour to feet per second for easier calculations. Knowing 1 mile is 5280 feet and 1 hour is 3600 seconds, \[ 45 \text{ mi/h} = \frac{45 \times 5280}{3600} \text{ ft/s} \approx 66 \text{ ft/s}. \]
03

Find the Deceleration

Use the formula for frictional force: \[ F_{\text{friction}} = \mu_k \cdot m \cdot g, \]where \( \mu_k = 0.750 \) is the coefficient of kinetic friction, \( m \) is the mass of the car, and \( g = 32.2 \text{ ft/s}^2 \) is the acceleration due to gravity. The deceleration (a) is equal to the frictional force per unit mass, \[ a = \mu_k \cdot g = 0.750 \times 32.2 \approx 24.15 \text{ ft/s}^2. \]
04

Use the Kinematic Equation

Apply the kinematic equation \[ v^2 = u^2 + 2as, \]where \( v = 0 \) (as the car stops), \( u \) is the initial velocity, \( a = -24.15 \text{ ft/s}^2 \) (negative due to deceleration), and \( s = 192 \text{ ft} \). Rearrange to find \( u \):\[ 0 = u^2 - 2 \times 24.15 \times 192. \]Solving for \( u \) gives \[ u = \sqrt{2 \times 24.15 \times 192} \approx 96 \text{ ft/s}. \]
05

Compare Initial Speed to Speed Limit

Convert \( u \) from feet per second to miles per hour to compare it with the speed limit:\[ u = \frac{96 \times 3600}{5280} \text{ mi/h} \approx 65.45 \text{ mi/h}. \]Since the initial speed, approximately 65.45 mi/h, is greater than the speed limit of 45 mi/h, the driver was speeding.
06

Conclusion

The driver's initial speed of about 65.45 mi/h was above the speed limit of 45 mi/h, indicating that he was indeed speeding.

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Key Concepts

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

Frictional Force
Frictional force is a crucial player in understanding car skidding. It refers to the force resisting the relative motion of a sliding object on a surface. In this context, the frictional force acts between the car's tires and the road as the car tries to stop.

There are different types of friction, but here we focus on kinetic friction because the car's wheels are locked and skidding across the pavement. The kinetic friction is what slows the car down, acting as a force opposite to the car's initial motion.

It’s determined by the formula: \[F_{\text{friction}} = \mu_k \times m \times g, \]where
  • \(F_{\text{friction}}\) is the frictional force,
  • \(\mu_k\) is the coefficient of kinetic friction,
  • \(m\) is the mass of the car, and
  • \(g\) is the acceleration due to gravity.
The frictional force is what causes the car to decelerate and eventually stop. Understanding this helps in determining whether a car was speeding by analyzing the length of the skid marks.
Coefficient of Kinetic Friction
The coefficient of kinetic friction, denoted as \(\mu_k\), characterizes how much frictional force exists between the two surfaces in contact during motion. It is a dimensionless quantity representing the ratio between the force of kinetic friction and the normal force acting on the object.

In our case, \(\mu_k = 0.750\) suggests relatively high friction between the road and the car's tires. It implies that the road and tires form a surface interaction that effectively slows down the car once the brakes are applied and the tires locked.

To determine the car's speed when it started skidding, the coefficient helps calculate the frictional force as \[a = \mu_k \times g\]where:
  • \(a\) is the acceleration (or deceleration in this scenario).
Knowing \(\mu_k\) leads to finding the deceleration, a critical step in checking how fast the vehicle was traveling initially.
Constant Acceleration
Constant acceleration means that the rate of change of velocity remains the same over time. In this scenario, the term is part of the narrative because when the car's brakes are applied, it experiences a steady deceleration.

Understanding constant acceleration is fundamental in using kinematic equations to solve motion problems. The crucial equation in this context is: \[v^2 = u^2 + 2as,\]where:
  • \(v\) is the final velocity (0 ft/s here because the car stops),
  • \(u\) is the initial velocity,
  • \(a\) is the constant acceleration (negative because it's deceleration), and
  • \(s\) is the displacement (192 ft in this case).
This equation allows us to discover how fast the car was going before it began stopping. With constant acceleration, calculations become straightforward as the equations do not change their form, simplifying the computational process within physics problems.

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Most popular questions from this chapter

Two objects, with masses 5.00 kg and 2.00 kg, hang 0.600 m above the floor from the ends of a cord that is 6.00 m long and passes over a frictionless pulley. Both objects start from rest. Find the maximum height reached by the 2.00-kg object.

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