Tag Archives: College Physics Solutions

College Physics by Openstax Chapter 2 Problem 10

Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of approximately 4 cm/year. Assuming this to be a constant rate, how many years will pass before the radius of the Moon’s orbit increases by 3.84×106 m(1%)?

Solution:

From the formula v‾=ΔxΔt\overline{v}=\frac{\Delta x}{\Delta t}, we can solve for Δt \Delta t as follows

Δt=Δxv‾=3.84×106 m4 cm/year×100 cm1 m=96 000 000 years=96.0×106 years  (Answer)\begin{align*} \Delta \text{t} & = \frac{\Delta x}{\overline{v}} \\ & = \frac{3.84\times 10^6\:\text{m}}{4\:\text{cm/year}}\times \frac{100\:\text{cm}}{1\:\text{m}} \\ & =96\:000\:000\:\text{years} \\ & =96.0\times 10^6\:\text{years} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

It will take about 96 million years.

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College Physics by Openstax Chapter 2 Problem 11

A student drove to the university from her home and noted that the odometer reading of her car increased by 12.0 km. The trip took 18.0 min.

(a) What was her average speed?

(b) If the straight-line distance from her home to the university is 10.3 km in a direction 25° S of E, what was her average velocity?

(c) If she returned home by the same path 7 h 30 min after she left, what were her average speed and velocity for the entire trip?

Solution:

Part A

The average speed is 

speed=distancetime=12 km18 mins×60 mins1 hr=40 km/hr  (Answer)\begin{align*} \text{speed} & = \frac{\text{distance}}{\text{time}}\\ &= \frac{12\:\text{km}}{18\:\text{mins}}\times \frac{60\:\text{mins}}{1\:\text{hr}} \\ & =40\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The average velocity is

v‾=Δx Δt=10.3 km18.0 min⁡ ×60 mins1 hr=34.33 km/hr  (Answer)\begin{align*} \overline{v} & =\frac{\Delta \text{x}\:}{\Delta \text{t}} \\ & =\frac{10.3\:\text{km}}{18.0\:\min \:}\times \frac{60\:\text{mins}}{1\:\text{hr}} \\ &=34.33\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The direction of the velocity is 25° S of E   (Answer)\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right).

Part C

The average speed is 

speed=distancetime=12.0 km×27.5 hr=3.2 km/hr  (Answer)\begin{align*} \text{speed} & = \frac{\text{distance}}{\text{time}}\\ & =\frac{12.0\:\text{km}\times 2}{7.5\:\text{hr}} \\ & =3.2\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

And the average velocity is

v‾=0  (Answer)\begin{align*} \overline{v}=0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The average velocity is zero since the total displacement is zero.

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College Physics by Openstax Chapter 2 Problem 9

On May 26, 1934, a streamlined, stainless steel diesel train called the Zephyr set the world’s nonstop long-distance speed record for trains. Its run from Denver to Chicago took 13 hours, 4 minutes, 58 seconds, and was witnessed by more than a million people along the route. The total distance traveled was 1633.8 km. What was its average speed in km/h and m/s?

Solution:

The total time of travel is converted to seconds.

t=(13 h ×3600 s1 hr)+(4 mins ×60 s1 min⁡)+58 sec⁡t=47 098 seconds\begin{align*} \text{t} & =\left(13\:\text{h}\:\times \frac{3600\:\text{s}}{1\:\text{hr}}\right)+\left(4\:\text{mins}\:\times \frac{60\:\text{s}}{1\:\min }\right)+58\:\sec \\ \text{t} & =47\:098\:\text{seconds} \end{align*}

The total time of travel in hours

 t=(47 098 seconds)(1 h3600 sec⁡)=13.0828 hours \text{t}=\left(47\:098\:\text{seconds}\right)\left(\frac{1\:\text{h}}{3600\:\sec }\right)=13.0828\:\text{hours}

Therefore, the average speed in km/hr is

speed in km/hr=distance traveledtime=1633.8 km13.0828 hr=124.88 km/hr  (Answer)\begin{align*} \text{speed in km/hr} & =\frac{\text{distance traveled}}{\text{time}} \\ & =\frac{1633.8\:\text{km}}{13.0828\:\text{hr}} \\ & =124.88\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

And the average speed in m/s is

speed in m/s=1 633 800 m47 098 s=34.689 m/s  (Answer)\begin{align*} \text{speed in m/s} & =\frac{1\:633\:800\:\text{m}}{47\:098\:\text{s}} \\ & =34.689\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}
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Problem 2-8: Motion of land mass around the San Andreas fault

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PROBLEM:

Land west of the San Andreas fault in southern California is moving at an average velocity of about 6 cm/y northwest relative to land east of the fault. Los Angeles is west of the fault and may thus someday be at the same latitude as San Francisco, which is east of the fault. How far in the future will this occur if the displacement to be made is 590 km northwest, assuming the motion remains constant?

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SOLUTION:

From the formula v‾=ΔxΔt\overline{v}=\frac{\Delta x}{\Delta t}, we can solve for Δt\Delta t as follows

Δt=Δxv‾=5.90×105 m6 cm/year×100 cm1 m=9.83×106 years  (Answer)\begin{align*} \Delta t & =\frac{\Delta x}{\overline{v}} \\ \\ & =\frac{5.90 \times 10^{5}\ \text{m}}{6\ \text{cm/year}}\times \frac{100\ \text{cm}}{1\ \text{m}} \\ \\ & = 9.83 \times 10^{6}\ \text{years} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

Therefore, it will take about 9.83 million years.

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Problem 2-7: The rate at which the North American and European continents are moving apart

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PROBLEM:

The North American and European continents are moving apart at a rate of about 3 cm/y. At this rate how long will it take them to drift 500 km farther apart than they are at present?

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SOLUTION:

We know that the formula for velocity is

velocity=displacementtime\text{velocity}=\frac{\text{displacement}}{\text{time}}

If we rearrange the formula and solve for time, we come up with

time=displacementvelocity\text{time}=\frac{\text{displacement}}{\text{velocity}}

Substituting the given values,

Δt=500 km3 cm/year=5×105 m3 cm/year×100 cm1 m=1.67×107years  (Answer)\begin{align*} \Delta \text{t} & =\frac{500\:\text{km}}{3\:\text{cm/year}} \\ \\ & =\frac{5\times 10^5\text{ m}}{3\:\text{cm/year}}\times \frac{100\:\text{cm}}{1\:\text{m}} \\ \\ & =1.67\times 10^7\text{years} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Therefore, the time it will take is about 1.67×107 years 1.67\times 10^{7}\ \text{years}.

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Problem 2-6: The average speed and average velocity of a spinning helicopter blade

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PROBLEM:

A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation.

(a) Calculate the average speed of the blade tip in the helicopter’s frame of reference.

(b) What is its average velocity over one revolution?

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SOLUTION:

Part A

The average speed of the blade tip is equal to the distance traveled divided by the time elapsed.

speed=distance traveledtime elapsed=2πrt=2π(5.00m)60 s×100 rev=52.36 m/s  (Answer)\begin{align*} \text{speed} & =\frac{ \text{distance traveled}}{ \text{time elapsed}} \\ \\ & =\frac{2\pi \text{r}}{ \text{t}} \\ \\ & =\frac{2\pi \left(5.00\text{m}\right)}{60\:\text{s}}\times 100\:\text{rev} \\ \\ & =52.36\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

After one revolution, the tip of the blade is at the same position as it is originally. This means that the displacement is zero. Thus, the velocity is zero.

v=0 m/s  (Answer)\text{v}=0\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 2-5: The Earth’s average speed and velocity

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PROBLEM:

(a) Calculate Earth’s average speed relative to the Sun.

(b) What is its average velocity over a period of one year?

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SOLUTION:

Part A

The average speed of the earth is equal to the distance traveled divided by time.

Average speed of the Earth=Distance TraveledTotal time of Travel=2πrt=2π(1.50×1011 m)365.25 days×1 day24 hours×1 hour3600 sec=2.99×104 m/s  (Answer)\begin{align*} \text{Average speed of the Earth} & =\frac{\text{Distance Traveled}}{\text{Total time of Travel}} \\ \\ & =\frac{2\pi \text{r}}{\text{t}} \\ \\ & =\frac{2\pi \left(1.50\times 10^{11}\text{ m}\right)}{365.25\:\text{days}}\times \frac{1\:\text{day}}{24\:\text{hours}}\times \frac{1\:\text{hour}}{3600\:\text{sec}} \\ \\ & =2.99\times 10^4\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

After a period of 1 year, the planet Earth has already returned to its original position with respect to the Sun. This means that we do not have any displacement. Therefore, the average velocity is zero.

Average velocity=0 m/s  (Answer)\text{Average velocity}=0\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 2-4: Distance and Displacement of a given Path D

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PROBLEM:

Find the following for path D in the figure:

(a) The distance traveled.

(b) The magnitude of the displacement from start to finish.

(c) The displacement from start to finish. 

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SOLUTION:

Part A

The position of D started from 9 m, then went to 3 m, and then went back to 5 m. The distance traveled by D is the sum of all the paths

distance=6 m+2 m=8 m  (Answer)\text{distance}=6\:\text{m}+2\:\text{m}=8\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

The magnitude of the displacement is the difference between the final position and the initial position WITHOUT regard to sign

∣Δx∣=∣9 m−5 m∣=4 m  (Answer)\left|\Delta x\right|=\left|9\:\text{m}-5\:\text{m}\right|=4\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part C

The displacement is the difference between the final position and the initial position TAKING INTO ACCOUNT the sign

Δx=5 m−9 m=−4 m  (Answer)\Delta x=5\:\text{m}-9\:\text{m}=-4\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 2-3: Distance and Displacement of a given Path C

PROBLEM:

Find the following for path C in the figure:

(a) The distance traveled.

(b) The magnitude of the displacement from start to finish.

(c) The displacement from start to finish.

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SOLUTION:

Part a

The distance is the sum of all the paths of C.

distance=8 m+2 m+2 m=12 m  (Answer)\text{distance}=8\:\text{m}+2\:\text{m}+2\:\text{m}=12\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part b

The magnitude of the displacement is the difference between the final position and the initial position without regard to the sign.

∣Δx∣=∣10 m−2 m∣=8 m  (Answer)\left|\Delta x\right|=\left|10\:\text{m}-2\:\text{m}\right|=8\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part c

The displacement is the difference between the final position and the initial position, taking into account the sign

Δx=10 m−2 m=+8 m  (Answer)\Delta x=10\:\text{m}-2\:\text{m}=+8\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 2-2: Distance and displacement of a given Path B

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PROBLEM:

Find the following for path B in the figure:

(a) The distance traveled.

(b) The magnitude of the displacement from start to finish.

(c) The displacement from start to finish.

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SOLUTION:

Part a

Based on the figure, B travels from 12 to 7. The distance traveled is 5 meters.

Part b

The magnitude of the displacement is 5 meters.

Part c

The displacement is calculated keeping in mind the sign. The motion started at 12 and ended at 7. Therefore, the displacement is

Δx=7 m−12 m=−5 meters  (Answer)\Delta x=7\:\text{m}-12\:\text{m}=-5\:\text{meters} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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