/*! 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 22 A car is traveling along a strai... [FREE SOLUTION] | 91Ó°ÊÓ

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A car is traveling along a straight road at a velocity of \(+36.0 \mathrm{m} / \mathrm{s}\) when its engine cuts out. For the next twelve seconds the car slows down, and its average acceleration is \(\bar{a}_{1}\). For the next six seconds the car slows down further, and its average acceleration is \(\bar{a}_{2 \cdot}\) The velocity of the car at the end of the eighteen-second period is \(+28.0 \mathrm{m} / \mathrm{s} .\) The ratio of the average acceleration values is \(\bar{a}_{1} / \bar{a}_{2}=1.50 .\) Find the velocity of the car at the end of the initial twelve-second interval.

Short Answer

Expert verified
The velocity at the end of the first twelve-second interval is 30.0 m/s.

Step by step solution

01

Understand the motion segments

Divide the problem into two segments of deceleration: the first 12-second segment with acceleration \(\bar{a}_1\) and the second 6-second segment with acceleration \(\bar{a}_2\). We have information about the total change in velocity over the entire 18-second period.
02

Use the average acceleration formulas

For each time segment, use the formula for average acceleration: \(\bar{a} = \frac{\Delta v}{\Delta t}\).1. For the first segment: \(\bar{a}_1 = \frac{v_{12} - v_0}{12}\), where \(v_0 = 36.0\,\mathrm{m/s}\) is the initial velocity and \(v_{12}\) is the velocity at the end of the 12 seconds.2. For the second segment: \(\bar{a}_2 = \frac{28.0 - v_{12}}{6}\).
03

Use the given acceleration ratio

We know \(\frac{\bar{a}_1}{\bar{a}_2} = 1.50\). Substitute the expressions of \(\bar{a}_1\) and \(\bar{a}_2\) from Step 2 into this ratio:\[\frac{\frac{v_{12} - 36.0}{12}}{\frac{28.0 - v_{12}}{6}} = 1.50\]
04

Simplify and solve the equation

Cross-multiply to clear the fractions:\[(v_{12} - 36.0) \times 6 = 1.50 \times (28.0 - v_{12}) \times 12\]Expand and simplify:\[6v_{12} - 216.0 = 18.0(28.0 - v_{12})\]\[6v_{12} - 216.0 = 504.0 - 18.0v_{12}\]Combine like terms:\[24v_{12} = 720.0\]\[v_{12} = 30.0\, \mathrm{m/s}\]
05

Verify the solution

Substitute \(v_{12} = 30.0\, \mathrm{m/s}\) back into the equations for \(\bar{a}_1\) and \(\bar{a}_2\) to verify the solution. If both accelerations satisfy the given acceleration ratio, the result is confirmed.

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Key Concepts

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

Understanding Average Acceleration
When solving physics problems involving motion, the term "average acceleration" becomes quite significant. Average acceleration is basically the rate at which velocity changes over a certain period of time. It's determined by dividing the change in velocity (\(\Delta v\)) by the time interval (\(\Delta t\)) over which this change occurs. In simpler terms, it's how quickly something speeds up or slows down on average. This is important when an object's speed is not constant over time.
Let's illustrate this with the exercise example where a car slows down as its engine cuts out. Initially, its average acceleration for the first 12 seconds is labeled \(\bar{a}_1\). This marks how quickly its speed decreases in this period. By calculating \(\bar{a}_1\), we can understand how much the car slows on average during these seconds. The formula to find this is:
  • \(\bar{a}_1 = \frac{v_{12} - v_0}{12}\)
where \(v_0\) is the initial velocity and \(v_{12}\) is the velocity after 12 seconds.
Exploring Deceleration
Deceleration is just another way of describing negative acceleration, where an object slows down. As acceleration turns negative, it indicates a decrease in velocity. In the problem we've looked at, the entire 18-second scenario is all about understanding deceleration.
For the second phase, lasting 6 seconds, the car keeps slowing down but likely at a different rate, noted by \(\bar{a}_2\). Finding \(\bar{a}_2\) tells us about the continuation of the deceleration. The formula used for this period of time is:
  • \(\bar{a}_2 = \frac{28.0 - v_{12}}{6}\)

This formula will help you figure out how much more the car decelerates in the next 6 seconds. In essence, any time you encounter deceleration in physics, it's crucial to note how the object's speed reduces over time.
Mastering Velocity Calculation
To master physics problems like the one we're dissecting, understanding how to calculate velocity is a must. Velocity isn't just speed; it's speed in a specific direction. Here, the car initially travels at +36.0 m/s, indicating both magnitude and direction.
The challenge is to find out the car's velocity at the end of a 12-second interval before entering another phase of slowing down. Here's where you integrate the calculation skills to piece together how velocity changes because of different periods of acceleration. By applying the average acceleration formulas, we find the velocity at 12 seconds:
  • \(v_{12} = 30.0 \, \text{m/s}\)

This shows us how fast the car was moving just before it entered the next acceleration segment. Remember, solving such problems involves systematically applying your understanding of kinematics to track how the object's motion evolves over time.

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Most popular questions from this chapter

A hot-air balloon is rising upward with a constant speed of \(2.50 \mathrm{m} / \mathrm{s}\). When the balloon is \(3.00 \mathrm{m}\) above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?

A locomotive is accelerating at \(1.6 \mathrm{m} / \mathrm{s}^{2} .\) It passes through a \(20.0-\mathrm{m}-\) wide erossing in a time of \(2.4 \mathrm{s}\). After the locomotive leaves the crossing, how much time is required until its speed reaches \(32 \mathrm{m} / \mathrm{s} ?\)

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 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?

A woman on a bridge \(75.0 \mathrm{m}\) high sees a raft floating at a constant speed on the river below. Trying to hit the raft, she drops a stone from rest when the raft has \(7.00 \mathrm{m}\) more to travel before passing under the bridge. The stone hits the water \(4.00 \mathrm{m}\) in front of the raft. Find the speed of the raft.

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