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Chapter 2: Problem 13

After you apply the brakes, the acceleration of your car is in the opposite direction to its velocity. If the acceleration of your car remains constant, describe the motion of your car.

Short Answer

Expert verified
Question: Describe the motion of a car when its acceleration is constant and in the opposite direction of its velocity after applying brakes. Answer: When a car's acceleration is constant and in the opposite direction of its velocity after applying brakes, this causes the car to decelerate. The constant deceleration results in a gradual reduction in the car's speed, leading it to eventually come to a stop. During deceleration, the car's displacement will continue to increase but at a decreasing rate.

Step by step solution

01

Understand the given scenario

In this situation, the car is initially moving in a certain direction with a positive velocity. When the brakes are applied, the acceleration is in the opposite direction, which means that it is negative.
02

Understand the concept of deceleration and how it affects the car's motion

Deceleration is the process of slowing down an object in motion. It is represented by a negative acceleration. In this case, the car's acceleration is in the opposite direction to its velocity, which is deceleration.
03

Describe the effect of the constant deceleration on velocity

As the car is constantly decelerating, the velocity will decrease over time. This will manifest as a gradual reduction in the car's speed until it eventually comes to a stop.
04

Explain the motion of the car using the equations of motion

We can use the equations of motion to describe the car's deceleration more mathematically. One such equation is: v = u + at Here, v is the final velocity, u is the initial velocity, a is the acceleration (deceleration in our case), and t is the time taken. Since the acceleration is constant and negative, the car's velocity will decrease linearly over time.
05

Describe the motion of the car in terms of displacement

As the car decelerates, it will continue to move forward and its displacement will increase over time, but at a decreasing rate. The displacement can be represented by the following equation: s = ut + (1/2)at^2 In this equation, s is the displacement, u is the initial velocity, a is the acceleration, and t is the time taken. Since the acceleration is negative, the second term of the equation will decrease over time, causing the rate of increase in the car's displacement to decrease as well. This means that the car's displacement will increase at a lower rate until it eventually comes to a stop.

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Most popular questions from this chapter

A car is traveling due west at \(20.0 \mathrm{~m} / \mathrm{s}\). Find the velocity of the car after \(3.00 \mathrm{~s}\) if its acceleration is \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) due west. Assume the acceleration remains constant. a) \(17.0 \mathrm{~m} / \mathrm{s}\) west b) \(17.0 \mathrm{~m} / \mathrm{s}\) east c) \(23.0 \mathrm{~m} / \mathrm{s}\) west d) \(23.0 \mathrm{~m} / \mathrm{s}\) east e) \(11.0 \mathrm{~m} / \mathrm{s}\) south

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The vertical position of a ball suspended by a rubber band is given by the equation $$ y(t)=(3.8 \mathrm{~m}) \sin (0.46 t / \mathrm{s}-0.31)-(0.2 \mathrm{~m} / \mathrm{s}) t+5.0 \mathrm{~m} $$ a) What are the equations for velocity and acceleration for this ball? b) For what times between 0 and \(30 \mathrm{~s}\) is the acceleration zero?

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