/*! 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 17 A motorcycle has a constant acce... [FREE SOLUTION] | 91Ó°ÊÓ

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A motorcycle has a constant acceleration of 2.5 \(\mathrm{m} / \mathrm{s}^{2} .\) Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from (a) 21 to \(31 \mathrm{m} / \mathrm{s},\) and \(\quad\) (b) 51 to 61 \(\mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
4 seconds for both scenarios.

Step by step solution

01

Understand the Problem

We are given a motorcycle with a constant acceleration of 2.5 m/s². The task is to find the time required for speed changes in two scenarios: from 21 m/s to 31 m/s and from 51 m/s to 61 m/s.
02

Use the Formula for Time with Constant Acceleration

The formula for time required to change velocity with constant acceleration is: \( t = \frac{\Delta v}{a} \), where \( \Delta v \) is the change in velocity, and \( a \) is the acceleration.
03

Calculate Time for First Scenario

For the first scenario, \( \Delta v = 31 \mathrm{m/s} - 21 \mathrm{m/s} = 10 \mathrm{m/s} \). The time is \( t = \frac{10 \mathrm{m/s}}{2.5 \mathrm{m/s}^2} = 4 \mathrm{s} \).
04

Calculate Time for Second Scenario

For the second scenario, \( \Delta v = 61 \mathrm{m/s} - 51 \mathrm{m/s} = 10 \mathrm{m/s} \). The time is \( t = \frac{10 \mathrm{m/s}}{2.5 \mathrm{m/s}^2} = 4 \mathrm{s} \).
05

Conclusion

In both scenarios, the time required for the motorcycle to change its speed is 4 seconds due to symmetry in speed change and constant acceleration.

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Key Concepts

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

Constant Acceleration
Imagine you're riding a motorcycle that keeps gaining speed at a steady rate. This is what we call constant acceleration. It means that the change in speed occurs uniformly over time. For the motorcycle in our exercise, it has a constant acceleration of 2.5 \(\mathrm{m/s^2}\). This value tells us how much the speed increases every second. So, with constant acceleration, if we start at rest, after the first second, the speed might be 2.5 m/s, after the second second, 5 m/s, and so on. Constant acceleration makes it easy to predict how speed changes over time, which is why the concept is so important when solving our problem.
Velocity
Velocity is a term that describes how fast and in which direction something is moving. In this exercise, we're focusing on velocity in the context of the motorcycle's movement. Velocity is more than just speed because it includes direction. However, for this particular problem, both the velocity and acceleration point in the same direction, making our calculations a bit simpler. In the scenarios provided, we're interested in the motorcycle changing its velocity from 21 m/s to 31 m/s and from 51 m/s to 61 m/s. These changes in velocity help us to understand exactly how fast the motorcycle is traveling at different instances and how long these transitions take.
Time Calculation
To find out how much time it takes for the motorcycle to change its speed, we use a simple formula derived from the basics of physics. The formula is: \( t = \frac{\Delta v}{a} \). Here, \( t\) is the time, \( \Delta v \) is the change in velocity, and \( a \) is the acceleration. This formula tells us that the time required for a velocity change is directly related to both the amount of acceleration and how much the velocity changes. By knowing the change in speed (\( \Delta v \)= 10 m/s) and constant acceleration (2.5 m/s²), we've calculated that the motorcycle's speed changes in 4 seconds for both scenarios. This calculation shows the straightforward relationship between these three important variables.
Change in Speed
The change in speed, or change in velocity, is a key concept in understanding motion. In our exercise, we focus on how the speed shifts from one value to another due to acceleration. Change in speed is calculated by finding the difference between the final velocity and the initial velocity. In both scenarios presented, the change is consistent, at 10 m/s, demonstrating how acceleration impacts speed uniformly over time. Understanding this change is essential for predicting motion and solving problems involving moving objects. Knowing how speed changes helps us make accurate estimations about how objects travel under constant acceleration, a common real-life scenario in kinematics.

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Most popular questions from this chapter

A race driver has made a pit stop to refuel. After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is \(6.0 \mathrm{m} / \mathrm{s}^{2} ;\) after 4.0 \(\mathrm{s}\) he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant velocity of 70.0 \(\mathrm{m} / \mathrm{s}\) overtakes and passes the entering car. The entering car maintains its acceleration. How much time is required for the entering car to catch the other car?

A hot-air balloon is rising straight up at a constant speed of 7.0 \(\mathrm{m} / \mathrm{s}\) . When the balloon is 12.0 \(\mathrm{m}\) above the ground, a gun fires a pellet straight up from ground level with an initial speed of 30.0 \(\mathrm{m} / \mathrm{s}\) . Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above ground are these places?

(a) What is the magnitude of the average acceleration of a skier who, starting from reaches a speed of 8.0 \(\mathrm{m} / \mathrm{s}\) when going down a slope for 5.0 \(\mathrm{s} ? \quad\) (b) How far does the skier travel in this time?

A jetliner, traveling northward, is landing with a speed of 69 \(\mathrm{m} / \mathrm{s}\) . Once the jet touches down, it has 750 \(\mathrm{m}\) of runway in which to reduce its speed to 6.1 \(\mathrm{m} / \mathrm{s}\) . Compute the average acceleration (magnitude and direction) of the plane during landing.

In a quarter-mile drag race, two cars start simultaneously from rest, and each accelerates at a constant rate until it either reaches its maximum speed or crosses the finish line. Car A has an acceleration of 11.0 \(\mathrm{m} / \mathrm{s}^{2}\) and a maximum speed of 106 \(\mathrm{m} / \mathrm{s}\) . Car \(\mathrm{B}\) has an acceleration of 11.6 \(\mathrm{m} / \mathrm{s}^{2}\) and a maximum speed of 92.4 \(\mathrm{m} / \mathrm{s}\) . Which car wins the race, and by how many seconds?

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