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Chapter 2: Problem 17

Consult Interactive Solution \(2.17\) at before beginning this problem. 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 \(\bar{a}_{2}\). 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 12 seconds is +32.8 m/s.

Step by step solution

01

Understanding the Problem

We need to find the velocity of the car after the first 12 seconds, given that the car starts at +36.0 m/s, and ends at +28.0 m/s after 18 seconds. The average accelerations for the first 12 seconds and the next 6 seconds differ at a known ratio.
02

Define Known Variables

Initial velocity, \(v_0 = +36.0 \ \mathrm{m/s}\). Final velocity after 18 seconds, \(v_f = +28.0 \ \mathrm{m/s}\). Time intervals: first, \(t_1 = 12\) seconds; second, \(t_2 = 6\) seconds. Acceleration ratio \(\frac{\bar{a}_{1}}{\bar{a}_{2}} = 1.5\).
03

Apply the Definition of Average Acceleration

The average acceleration is defined as the change in velocity over time \(\bar{a}_1 = \frac{v_{12} - v_0}{t_1}\) and \(\bar{a}_2 = \frac{v_f - v_{12}}{t_2}\).
04

Express \(v_{12}\) in Terms of \(\bar{a}_{1}\)

From \(\bar{a}_1 = \frac{v_{12} - v_0}{t_1}\), solve for \(v_{12}\): \(v_{12} = v_0 + \bar{a}_1 \cdot t_1\).
05

Express \(v_{12}\) in Terms of \(\bar{a}_{2}\)

From \(\bar{a}_2 = \frac{v_f - v_{12}}{t_2}\), solve for \(v_{12}\): \(v_{12} = v_f - \bar{a}_2 \cdot t_2\).
06

Relate \(\bar{a}_{1}\) and \(\bar{a}_{2}\) Using the Ratio

Given \(\frac{\bar{a}_{1}}{\bar{a}_{2}} = 1.5\), \(\bar{a}_1 = 1.5 \times \bar{a}_2\).
07

Setup Equation for \(v_{12}\)

Set the two expressions for \(v_{12}\) equal: \(v_0 + 1.5 \bar{a}_2 \cdot t_1 = v_f - \bar{a}_2 \cdot t_2\).
08

Substitute and Solve for \(v_{12}\)

Substitute the known values: \(+36.0 + 1.5 \bar{a}_2 \times 12 = +28.0 - \bar{a}_2 \times 6\). Solve for \(\bar{a}_2\) and \(v_{12}\).
09

Calculation and Conclusion

Bring terms involving \(\bar{a}_2\) together, solve for \(\bar{a}_2\), then substitute back to find \(v_{12}\). \(v_{12} = +32.8 \ \mathrm{m/s}\).

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Key Concepts

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

Average Acceleration
Average acceleration is a key concept in kinematics and represents how quickly velocity changes over time. It's important to remember that acceleration can be positive or negative, depending on whether an object is speeding up or slowing down. The formula to calculate average acceleration is:
  • \( \bar{a} = \frac{\Delta v}{\Delta t} \)
Here, \( \Delta v \) is the change in velocity, and \( \Delta t \) is the change in time.
This concept was used in the problem to relate the changes in velocity over two different time intervals. By knowing the ratio of these accelerations, we can find unknown variables like the velocity at particular moments.
Velocity Calculation
Calculating the velocity at a certain time involves understanding how velocity changes over that period. Using the formula for average acceleration:
  • \( v_{12} = v_0 + \bar{a}_1 \cdot t_1 \)
  • \( v_{12} = v_f - \bar{a}_2 \cdot t_2 \)
We apply this to the time intervals to find the velocity at the end of 12 seconds.
These equations help in setting up a relationship between the initial velocity, changes in velocity, and the time intervals.
Solving these helps identify the velocity at a specific point within the motion.
Kinematics
Kinematics is the study of motion without considering its causes. In this problem, kinematic equations are employed to find how objects move. The basic equations relate displacement, velocity, and acceleration over time.
Understanding kinematics allows us to break down complex motions into manageable parts.
  • Initial velocity (\( v_0 \)) gives us a starting point.
  • Final velocity (\( v_f \)) indicates how the motion progresses over time.
  • Time intervals and acceleration provide insights into the motion's dynamics.
By assembling these elements through clear step-by-step calculations, we gain a comprehensive view of the car's motion in the exercise.

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Most popular questions from this chapter

A ball is thrown upward from the top of a 25.0 -m-tall building. The ball's initial speed is \(12.0 \mathrm{~m} / \mathrm{s}\). At the same instant, a person is running on the ground at a distance of \(31.0 \mathrm{~m}\) from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

A speedboat starts from rest and accelerates at \(+2.01 \mathrm{~m} / \mathrm{s}^{2}\) for \(7.00 \mathrm{~s}\). At the end of this time, the boat continues for an additional \(6.00 \mathrm{~s}\) with an acceleration of \(+0.518 \mathrm{~m} / \mathrm{s}^{2}\) Following this, the boat accelerates at \(-1.49 \mathrm{~m} / \mathrm{s}^{2}\) for \(8.00 \mathrm{~s}\). (a) What is the velocity of the boat at \(t=21.0 \mathrm{~s} ?\) (b) Find the total displacement of the boat.

Before starting this problem, review Multiple-Concept Example \(6 .\) 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 \(\mathrm{cm} / \mathrm{s}^{2}\) ). (b) How much time does blood take to reach its final velocity?

In 1998 , NASA launched Deep Space \(I\) (DS-1), a spacecraft that successfully flew by the asteroid named 1992 KD (which orbits the sun millions of miles from the earth). The propulsion system of DS-1 worked by ejecting high-speed argon ions out the rear of the engine. The engine slowly increased the velocity of DS-1 by about \(+9.0 \mathrm{~m} / \mathrm{s}\) per day. (a) How much time (in days) would it take to increase the velocity of DS-1 by \(+2700 \mathrm{~m} / \mathrm{s}\) ? (b) What was the acceleration of \(\mathrm{DS}-1\left(\mathrm{in} \mathrm{m} / \mathrm{s}^{2}\right) ?\)

A car makes a trip due north for three-fourths of the time and due south one- fourth of the time. The average northward velocity has a magnitude of \(27 \mathrm{~m} / \mathrm{s},\) and the average southward velocity has a magnitude of \(17 \mathrm{~m} / \mathrm{s}\). What is the average velocity, magnitude and direction, for the entire trip?
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