Tag Archives: College Physics by Openstax

College Physics by Openstax Chapter 2 Problem 66

Figure 2.68 shows the position graph for a particle for 6 s. (a) Draw the corresponding Velocity vs. Time graph. (b) What is the acceleration between 0 s and 2 s? (c) What happens to the acceleration at exactly 2 s?

Solution:

Part A

The velocity of the particle is the slope of the position vs time graph. Since the position graph is composed of straight lines, we can say that the velocity is constant for several time ranges.

Time Range	Slope of the Position vs Time Graph
0 to 2 seconds	$=\frac{2-0}{2-0}=1\:\text{m/s}$
2 to 3 seconds	$=\frac{-3-2}{3-2}=\frac{-5}{1}=-5\:\text{m/s}$
3 to 5 seconds	$=0 \ \text{m/s}$
5 to 6 seconds	$=\frac{-2-\left(-3\right)}{6-5}=\frac{1}{1}=1\:\text{m/s}$

Based on the data in the table, we can draw the velocity diagram

Part B

Since the velocity is constant between 0 seconds and 2 seconds, we say that the acceleration is 0.

Part C

Since there is a sudden change in velocity at exactly 2 seconds in a very short amount of time, we say that the acceleration is undefined in this case.

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College Physics by Openstax Chapter 2 Problem 65

A graph of v(t) is shown for a world-class track sprinter in a 100-m race. (See Figure 2.67). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at t=5 s? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

Solution:

Part A

To find the average velocity over the straight line graph of the velocity vs time shown, we just need to locate the midpoint of the line. In this case, the average speed for the first 4 seconds is

v_{\text{ave}}=6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

Looking at the graph, the velocity at exactly 5 seconds is 12 m/s.

Part C

If we are given the velocity-time graph, we can solve for the acceleration by solving for the slope of the line.

Consider the line from time zero to time, t=4 seconds. The slope, or acceleration, is

a=\text{slope}=\frac{12\:\text{m/s}-0\:\text{m/s}}{4\:\text{s}}=3\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part D

For the first 4 seconds, the distance traveled is equal to the area under the curve.

\text{distance}=\frac{1}{2}\left(4\:\sec \right)\left(12\:\text{m/s}\right)=24\:\text{m}

So, the sprinter traveled a total of 24 meters in the first 4 seconds. He still needs to travel a distance of 76 meters to cover the total racing distance. At the constant rate of 12 m/s, he can run the remaining distance by

\text{t}=\frac{\text{distance}}{\text{velocity}}=\frac{76\:\text{m}}{12\:\text{m/s}}=6.3\:\sec

Therefore, the total time of the sprint is

\text{t}_{\text{total}}=4\:\sec +6.3\:\sec =10.3\:\sec \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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College Physics by Openstax Chapter 2 Problem 49

You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down?

Solution:

The known values are $a=-9.80\:\text{m/s}^2$ ; $v_o=15.0\:\text{m/s}$ ; $y=7.00\:\text{m}$

The applicable formula is.

y=v_ot+\frac{1}{2}at^2

Using this formula, we can solve it in terms of time, $t$ .

t=\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a}

Substituting the known values, we have

\begin{align*} t & =\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a} \\ t & =\frac{-15.0\:\text{m/s}\pm \sqrt{\left(15.0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(7.00\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\ t&=\frac{-15.0\:\text{m/s}\pm 9.37\:\text{m/s}}{-9.80\:\text{m/s}^2} \end{align*}

We have two values for time, $t$ . These two values represent the times when the ball passes the tree branch.

t_1=\frac{-15.0\:m/s+9.37\:m/s}{-9.80\:m/s^2}=0.57\:sec \\ t_2=\frac{-15.0\:m/s-9.37\:m/s}{-9.80\:m/s^2}=2.49\:sec

Therefore, the total time between passing the branch is the difference between 2.49 seconds and 0.57 seconds.

t_2-t_1=2.49 \ \text{s} - 0.57 \ \text{s}=1.92 \ \text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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$Solution Guides to College Physics Chapter 2 Banner$

Chapter 2: Kinematics

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Displacement

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Time, Velocity, and Speed

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Acceleration

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Motion Equations for Constant Acceleration in One Dimension

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Falling Objects

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Graphical Analysis of One-Dimensional Motion

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$Solution Guide to College Physics by Openstax Chapter 1 Banner$

Chapter 1: Introduction: The Nature of Science and Physics

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Physical Quantities and Units

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Accuracy, Precision, and Significant Figures

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Approximation

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$Solution Guides for College Physics by Openstax Banner$

College Physics by Openstax

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