Tag Archives: Physics PDF solutions

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?

position graph for a particle for 6 s.
Figure 2.68

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 RangeSlope 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

velocity vs time graph
velocity vs time graph

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?

A graph of  v(t)  is shown for a world-class track sprinter in a 100-m race.
Figure 2.67

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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Physics for Scientists and Engineers 3E by R. Knight, P1.33


Estimate the average speed with which the hair on your head grows. Give your answer in both m/s and µm/hour. Briefly describe how you arrived at this estimate.


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Physics for Scientists and Engineers 3E by R. Knight, P1.32


Estimate the average speed with which you go from home to campus via whatever mode of transportation you use most commonly. Give your answer in both mph and m/s. Briefly describe how you arrived at this estimate.


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Physics for Scientists and Engineers 3E by R. Knight, P1.31


Estimate the height of a telephone pole. Give your answer in both feet and meters. Briefly describe how you arrived at this estimate.


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Physics for Scientists and Engineers 3E by R. Knight, P1.30


Estimate (don’t measure!) the length of a typical car. Give your answer in both feet and meters. Briefly describe how you arrived at this estimate.


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Physics for Scientists and Engineers 3E by R. Knight, P1.29


Compute the following numbers, applying the significant figure rule adopted in this textbook.

a. \displaystyle 12.5^3

b. \displaystyle 12.5\times 5.21

c. \displaystyle \sqrt{12.5}-1.2

d. \displaystyle 12.5^{-1}


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Physics for Scientists and Engineers 3E by R. Knight, P1.28


Compute the following numbers, applying the significant figure rule adopted in this textbook.

a. \displaystyle 33.3\times 25.4

b. \displaystyle 33.3-25.4

c. \displaystyle \sqrt{33.3}

d. \displaystyle 333.3\div 25.4


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Physics for Scientists and Engineers 3E by R. Knight, P1.27


Using the approximate conversion factors in Table 1.5, convert the following to SI units to English units without using your calculator.

a. 30 cm

b. 25 m/s

c. 5 km

d. 0.5 cm


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Physics for Scientists and Engineers 3E by R. Knight, P1.26


Using the approximate conversion factors in Table 1.5, convert the following to SI units without using your calculator.

a. 20 ft

b. 60 mi

c. 60 mph

d. 8 in


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