/*! 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 33 A car is traveling at \(20.0 \ma... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 33

A car is traveling at \(20.0 \mathrm{m} / \mathrm{s},\) and the driver sees a traffic light turn red. After \(0.530 \mathrm{s}\) (the reaction time), the driver applies the brakes, and the car decelerates at \(7.00 \mathrm{m} / \mathrm{s}^{2} .\) What is the stopping distance of the car, as measured from the point where the driver first sees the red light?

Short Answer

Expert verified
The stopping distance is 39.17 meters.

Step by step solution

01

Calculate the distance covered during reaction time

First, we determine how far the car travels during the driver's reaction time of 0.530 seconds while maintaining a constant speed of 20.0 m/s. The distance, denoted as \( d_1 \), can be calculated using the formula \( d = vt \), where \( v \) is the speed and \( t \) is the time.\[ d_1 = 20.0 \, \text{m/s} \times 0.530 \, \text{s} = 10.6 \, \text{m} \]
02

Calculate the stopping distance after braking

Next, we calculate the distance covered from the point where the brakes are applied (after the reaction time) until the car stops. We use the equation for stopping distance under uniform deceleration: \( v_f^2 = v_i^2 + 2a d \), where \( v_f \) is the final velocity (0 m/s), \( v_i \) is the initial velocity (20.0 m/s), \( a \) is the acceleration (-7.00 m/s虏, as it's deceleration), and \( d \) is the distance we want to find.Rearrange the formula to solve for \( d_2 \):\[ d_2 = \frac{v_f^2 - v_i^2}{2a} = \frac{0 - (20.0)^2}{2(-7.00)} = \frac{-400}{-14} = 28.57 \, \text{m} \]
03

Calculate the total stopping distance

To find the total stopping distance, add the distance covered during the reaction time (\( d_1 \)) and the braking distance (\( d_2 \)).\[ \text{Total stopping distance} = d_1 + d_2 = 10.6 \, \text{m} + 28.57 \, \text{m} = 39.17 \, \text{m} \]

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Key Concepts

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

Reaction Time
When discussing vehicle stopping distances, reaction time is crucial. It refers to the time taken by a driver to respond to a stimulus; in this case, the red traffic light. The driver may need this brief period to realize the need to stop and then act by applying the brakes. In this real-world scenario, a reaction time of 0.530 seconds was noted, which is relatively typical for many individuals. During this reaction period, the car continues to travel at the initial speed. Understanding this concept is critical because it affects how far a car travels before the driver even starts to slow it down.
Uniform Deceleration
Uniform deceleration occurs when a vehicle slows down at a constant rate. In the exercise, once the driver applies the brakes, the car begins to decelerate uniformly. The deceleration is consistent, meaning the velocity of the car decreases by the same amount each second. Here, the deceleration is 7.00 m/s虏, indicating that every second the car's speed reduces by 7.00 m/s. This concept is a key part of calculating stopping distance after brakes are applied as it allows us to use equations of motion to determine other variables like distance traveled.
Stopping Distance
Stopping distance is the total distance a vehicle travels before it comes to a complete halt. It is critical in evaluating road safety and is the sum of two main parts: the distance traveled during the reaction time and the distance during actual braking. In this example, the stopping distance includes the 10.6 meters covered during reaction time before brakes are applied, and the 28.57 meters covered under deceleration. This gives a total stopping distance of 39.17 meters. Understanding stopping distance helps in planning safer travel speeds and provides insight into the importance of quick response times and effective braking.
Equations of Motion
Equations of motion are fundamental in solving problems involving uniform acceleration or deceleration. These equations relate distance, velocity, acceleration, and time.In the context of stopping a vehicle, they allow us to calculate stopping distance effectively. The key equation used in our example is:
  • \[ v_f^2 = v_i^2 + 2a d \]
This equation helps calculate the braking distance (28.57 meters in our problem), where:
  • \( v_f \) is the final velocity (0 m/s here, since the car stops)
  • \( v_i \) is the initial velocity (20.0 m/s)
  • \( a \) represents acceleration (or deceleration rate, -7.00 m/s虏 in this case)
  • \( d \) is the distance covered under deceleration
Grasping these equations is essential for those studying kinematics, as they form the basis for analyzing motion under constant acceleration or deceleration.

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Most popular questions from this chapter

The left ventricle of the heart accelerates blood from rest to a velocity of \(+26 \mathrm{cm} / \mathrm{s}\). (a) If the displacement of the blood during the acceleration is \(+2.0 \mathrm{cm},\) determine its acceleration (in \(\left.\mathrm{cm} / \mathrm{s}^{2}\right)\) (b) How much time does blood take to reach its final velocity?

A golf ball is dropped from rest from a height of \(9.50 \mathrm{m}\). It hits the pavement, then bounces back up, rising just \(5.70 \mathrm{m}\) before falling back down again. A boy then catches the ball on the way down when it is \(1.20 \mathrm{m}\) above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.

A police car is traveling at a velocity of \(18.0 \mathrm{m} / \mathrm{s}\) due north, when a car zooms by at a constant velocity of \(42.0 \mathrm{m} / \mathrm{s}\) due north. After a reaction time of \(0.800 \mathrm{s}\) the policeman begins to pursue the speeder with an acceleration of \(5.00 \mathrm{m} / \mathrm{s}^{2} .\) Including the reaction time, how long does it take for the police car to catch up with the speeder?

The greatest height reported for a jump into an airbag is \(99.4 \mathrm{m}\) by stuntman Dan Koko. In 1948 he jumped from rest from the top of the Vegas World Hotel and Casino. He struck the airbag at a speed of \(39 \mathrm{m} / \mathrm{s}(88 \mathrm{mi} / \mathrm{h}) .\) To assess the effects of air resistance, determine how fast he would have been traveling on impact had air resistance been absent.

A diver springs upward with an initial speed of \(1.8 \mathrm{m} / \mathrm{s}\) from a \(3.0-\mathrm{m}\) board. (a) Find the velocity with which he strikes the water. / Hint: When the diver reaches the water, his displacement is \(y=-3.0 \mathrm{m}\) (measured from the board), assuming that the dowmward direction is chosen as the negative direction./ (b) What is the highest point he reaches above the water?
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