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Chapter 2: Problem 15

(a) Suppose that a NASCAR race car is moving to the right with a constant velocity of \(+82 \mathrm{m} / \mathrm{s}\). What is the average acceleration of the car? (b) Twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed. What is the average acceleration of the car?

Short Answer

Expert verified
(a) 0 m/s虏; (b) -13.67 m/s虏

Step by step solution

01

Understanding Constant Velocity

To find the average acceleration, it's important to remember that acceleration is the change in velocity over time. When a car is moving at a constant velocity, it means there is no change in speed or direction. If the NASCAR car moves to the right with a constant velocity of \(+82 \, \mathrm{m/s}\), the change in velocity is zero. Thus, the average acceleration in this case is \(0\, \mathrm{m/s^2}\).
02

Analyzing the Change in Direction

For part (b), the car initially has a velocity of \(+82 \, \mathrm{m/s}\) and later moves in the opposite direction with the same speed of \(-82 \, \mathrm{m/s}\). This indicates a change in direction, which means a change in velocity. The initial velocity \(v_i\) is \(+82 \, \mathrm{m/s}\) and the final velocity \(v_f\) is \(-82 \, \mathrm{m/s}\).
03

Calculating Change in Velocity

The change in velocity \(\Delta v\) can be calculated using the formula: \(\Delta v = v_f - v_i\). Substituting the given values, \(\Delta v = -82 \, \mathrm{m/s} - (+82 \, \mathrm{m/s}) = -164 \, \mathrm{m/s}\). This shows the total change in velocity from \(+82 \, \mathrm{m/s}\) to \(-82 \, \mathrm{m/s}\).
04

Calculating Average Acceleration

The average acceleration \(a_{avg}\) is calculated using the formula: \(a_{avg} = \frac{\Delta v}{\Delta t}\). The time interval \(\Delta t\) is given as 12 seconds. So, \(a_{avg} = \frac{-164 \, \mathrm{m/s}}{12 \, \mathrm{s}} = -13.67 \, \mathrm{m/s^2}\). This negative sign indicates that the acceleration acts in the opposite direction to the initial movement.

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Key Concepts

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

Constant Velocity
When we talk about constant velocity, we're referring to an object's steady pace without any changes in speed or direction. Imagine you're driving a car on a straight highway at 82 meters per second. This steady speed is your constant velocity.
  • No Speed Change: The car maintains the same speed without speeding up or slowing down.
  • No Direction Change: The car continues in the same straight path.

If there is no change in either direction or speed, the acceleration is zero. This is why, for the NASCAR car traveling at +82 m/s with constant velocity, the average acceleration is 0 \( \mathrm{m/s^2} \). It simply means the car isn't accelerating or decelerating but moving steadily along.
Change in Velocity
Velocity isn't just about speed; it's also about direction. When an object changes its direction or speed, its velocity changes. Let's explore this a bit more.
  • Direction Change: Even if the speed stays the same, a change in direction means a change in velocity.
  • Speed Change: Increase or decrease in the rate of motion also constitutes a velocity change.

In our exercise, the NASCAR race car initially moves at \(+82 \mathrm{m/s}\) and later at \(-82 \mathrm{m/s}\). Although the speed remains 82 \(\mathrm{m/s}\), the direction changes. This shift from positive to negative indicates a velocity change.
Time Interval
The concept of time interval is crucial for understanding acceleration. It refers to the duration over which changes occur. In physics, measuring how fast something happens means knowing both the change and how long it took.
  • Start and End Point: Consider when the event starts and ends. This gives you your measurement period.
  • Duration in Seconds: In many physics problems, like ours, time is often expressed in seconds.

For example, in our scenario, the car takes 12 seconds to change from moving in one direction to the opposite. This time interval \(\Delta t\) is key for calculating average acceleration.
Direction Change
A change in direction is a fundamental aspect of velocity change. It's not just about how fast you go but where you're headed. Imagine making a U-turn while driving; even if you maintain your speed, the change in course modifies your velocity.
  • Reversing Course: Moving from one direction to its opposite, like our NASCAR car switching from \(+82 \mathrm{m/s}\) to \(-82 \mathrm{m/s}\).
  • Effects on Velocity: Although speed remains constant, direction change results in a different velocity.

When the car changes its direction halfway around the track but retains the same speed, this direction change impacts the calculations of average acceleration significantly. The total change in velocity is determined by both speed and directional switches.

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Most popular questions from this chapter

The leader of a bicycle race is traveling with a constant velocity of \(+11.10 \mathrm{m} / \mathrm{s}\) and is \(10.0 \mathrm{m}\) ahead of the second-place cyclist. The secondplace cyclist has a velocity of \(+9.50 \mathrm{m} / \mathrm{s}\) and an acceleration of \(+1.20 \mathrm{m} / \mathrm{s}^{2}\). How much time elapses before he catches the leader?

A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part, she rides for 22 minutes at an average speed of \(7.2 \mathrm{m} / \mathrm{s}\). During the second part, she rides for 36 minutes at an average speed of \(5.1 \mathrm{m} / \mathrm{s} .\) Finally, during the third part, she rides for 8.0 minutes at an average speed of \(13 \mathrm{m} / \mathrm{s}\). (a) How far has the bicyclist traveled during the entire trip? (b) What is her average velocity for the trip?

A jet is taking off from the deck of an aircraft carrier, as shown in the image. Starting from rest, the jet is catapulted with a constant acceleration of \(+31 \mathrm{m} / \mathrm{s}^{2}\) along a straight line and reaches a velocity of \(+62 \mathrm{m} / \mathrm{s}\). Find the displacement of the jet. (a) A jet is being launched from an aircraft carrier. (b) During the launch, a catapult accelerates the jet down the flight deck.

Two arrows are shot vertically upward. The second arrow is shot after the first one, but while the first is still on its way up. The initial speeds are such that both arrows reach their maximum heights at the same instant, although these heights are different. Suppose that the initial speed of the first arrow is \(25.0 \mathrm{m} / \mathrm{s}\) and that the second arrow is fired \(1.20 \mathrm{s}\) after the first. Determine the initial speed of the second arrow.

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of \(86.0 \mathrm{m} / \mathrm{s}^{2}\) for 1.70 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?
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