Author Archives: Engineering Math

College Physics by Openstax Chapter 2 Problem 19

Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in m/s2 and in multiples of g (9.80 m/s2) ?

Solution:

The formula for acceleration is 

a‾=ΔvΔt\overline{a}=\frac{\Delta v}{\Delta t}

Substituting the given values

a‾=vf−v0Δta‾=6.5×103 m/s−0 m/s60.0 seca‾=108.33 m/s2  (Answer)\begin{align*} \overline{a} & = \frac{v_f-v_0}{\Delta t} \\ \overline{a} & =\frac{6.5\times 10^3\:\text{m/s}-0\:\text{m/s}}{60.0\:\text{sec}}\\ \overline{a} & =108.33\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

This can be expressed in multiples of g

a‾=108.33 m/s29.80 m/s2a‾=11.05g  (Answer)\begin{align*} \overline{a} & = \frac{108.33\:\text{m/s}^2}{9.80\:\text{m/s}^2}\\ \overline{a} & =11.05\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, the average acceleration is 108.33 m/s2 and can be expressed as 11.05g.

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College Physics by Openstax Chapter 2 Problem 18

A commuter backs her car out of her garage with an acceleration of 1.40 m/s2 .

(a) How long does it take her to reach a speed of 2.00 m/s?

(b) If she then brakes to a stop in 0.800 s, what is her deceleration?

Solution:

Part A

The formula for acceleration is

a‾=ΔvΔt\overline{a}=\frac{\Delta v}{\Delta t}

If we rearrange the formula by solving for Δt\Delta t, in terms of velocity and acceleration, we come up with

Δt=Δva‾\Delta t=\frac{\Delta v}{\overline{a}}

Substituting the given values, we have

Δt=Δva‾Δt=vf−v0a‾Δt=2.00 m/s−0 m/s1.40 m/s2Δt=1.43 seconds  (Answer)\begin{align*} \Delta t & =\frac{\Delta v}{\overline{a}} \\ \Delta t & = \frac{v_f-v_0}{\overline{a}} \\ \Delta t & =\frac{2.00 \ \text{m/s}-0 \ \text{m/s}}{1.40 \ \text{m/s}^2} \\ \Delta t & =1.43 \ \text{seconds} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The formula for acceleration (deceleration) is

a‾=ΔvΔt\overline{a}=\frac{\Delta v}{\Delta t}

Then substituting all the given values, we have

a‾=vf−v0Δta‾=0 m/s−2 m/s0.8 m/s2a‾=−2.50 m/s2  (Answer)\begin{align*} \overline{a} & = \frac{v_f-v_0}{\Delta t} \\ \overline{a} & = \frac{0 \ \text{m/s}-2\ \text{m/s}}{0.8 \ \text{m/s}^2} \\ \overline{a} & = -2.50 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 17

Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his

(a) acceleration and

(b) deceleration.

Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity.

Solution:

Part A

The formula for acceleration is

a‾=ΔvΔta‾=vf−v0tf−t0\begin{align*} \overline{a} & =\frac{\Delta v}{\Delta t} \\ \overline{a} & = \frac{v_f-v_0}{t_f-t_0} \\ \end{align*}

Substituting the given values

a‾=282 m/s−0 m/s5.00 sec⁡a‾=56.4 m/s2  (Answer)\begin{align*} \overline{a} & =\frac{282\:\text{m/s}-0\:\text{m/s}}{5.00\:\sec } \\ \overline{a} & =56.4\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The deceleration is 

a‾=0 m/s−282 m/s1.40 sa‾=−201.43 m/s2  (Answer)\begin{align*} \overline{a} & =\frac{0\:\text{m/s}-282\:\text{m/s}}{1.40\:\text{s}} \\ \overline{a} & =-201.43\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

In expressing the computed values in terms of g, we just divide them by 9.80.

The acceleration is

a‾=56.49.80=5.76g  (Answer)\overline{a}=\frac{56.4}{9.80}=5.76\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

The deceleration is

a‾=201.439.80=20.55g  (Answer)\overline{a}=\frac{201.43}{9.80}=20.55\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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College Physics by Openstax Chapter 2 Problem 16

A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration?

Solution:

The formula for acceleration is 

a‾=change in velocitychange in timea‾=ΔvΔta‾=vf−v0tf−t0\begin{align*} \overline{a} & =\frac{\text{change in velocity}}{\text{change in time}}\\ \overline{a} & =\frac{\Delta \text{v}}{\Delta t} \\ \overline{a} & =\frac{v_f-v_0}{t_f-t_0} \\ \end{align*}

Substituting the given values

a‾=30.0 m/s−0 m/s7.00 s=4.29 m/s2  (Answer)\begin{align*} \overline{a} & =\frac{30.0\:\text{m/s}-0\:\text{m/s}}{7.00\:\text{s}} \\ & =4.29\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 15

The planetary model of the atom pictures electrons orbiting the atomic nucleus much as planets orbit the Sun. In this model, you can view hydrogen, the simplest atom, as having a single electron in a circular orbit 1.06×10-10 m in diameter.

(a) If the average speed of the electron in this orbit is known to be 2.20×106 m/s, calculate the number of revolutions per second it makes about the nucleus.

(b) What is the electron’s average velocity?

Solution:

Part A

The formula to be used is

average speed=distancetimer=dt\begin{align*} \text{average speed} & =\frac{\text{distance}}{\text{time}} \\ r & = \frac{d}{t} \end{align*}

Rearranging the formula–solving for the distance

d=r×t\begin{align*} d=r\times t \end{align*}

Substituting the given values for 1 second period

d=(2.20×106 m/s)(1 s)=2.20×106 meters\begin{align*} d & = \left(2.20\times 10^6\:\text{m/s}\right)\left(1\:\text{s}\right) \\ & =2.20\times 10^6\:\text{meters} \end{align*}

This is the total distance traveled in 1 sec.

With the given radius, the total distance traveled in 1 revolution is

1 revolution=2πr=πd=π(1.06×10−10m)\begin{align*} 1\:\text{revolution} & =2\pi \text{r} \\ & =\pi \text{d} \\ &=\pi \left(1.06\times 10^{-10}\text{m}\right) \end{align*}

Therefore, the total number of revolutions traveled in 1 second is

no. of revolutions=total distancedistance in 1 revolution=2.20×106π(1.06×10−10)=6.61×1015 revolutions  (Answer)\begin{align*} \text{no. of revolutions} & = \frac{\text{total distance}}{\text{distance in 1 revolution}} \\ & = \frac{2.20\times 10^6}{\pi \left(1.06\times 10^{-10}\right)} \\ & =6.61\times 10^{15} \ \text{revolutions} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

In one complete revolution, the electron will go back to its original position. Thus, there is no net displacement. Therefore,

v‾=ΔxΔtv‾=0 m/s  (Answer)\begin{align*} \overline{v} & =\frac{\Delta x}{\Delta t} \\ \overline{v} & =0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 14

A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity

(a) for each of the three intervals and

(b) for the entire motion.

Solution:

Part A

The average velocity for each interval is computed using the formula

v‾=ΔxΔt\overline{v}=\frac{\Delta x}{\Delta t}

For the first interval

v1‾=15.0 meters2.50 sec⁡=6.00 m/s  (Answer) \overline{v_1}=\frac{15.0\:\text{meters}}{2.50\:\sec }=6.00\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

For the second interval

v2‾=−3.00 meters1.75 sec⁡=−1.71 m/s  (Answer)\overline{v_2}=\frac{-3.00\:\text{meters}}{1.75\:\sec }=-1.71\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

For the third interval

v3‾=21.0 m5.20 sec=4.04 m/s  (Answer)\overline{v_3}=\frac{21.0\:\text{m}}{5.20\:\text{sec}}=4.04\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

For the entire motion, we need displacement from the beginning to the end.

v‾=ΔxΔt=15 m−3 m+21 m2.50 s+1.75 s+5.20 s=33 m9.45 s=3.49 m/s  (Answer)\begin{align*} \overline{v}& =\frac{\Delta x}{\Delta t} \\ & =\frac{15\:\text{m}-3\:\text{m}+21\:\text{m}}{2.50\:\text{s}+1.75\:\text{s}+5.20\:\text{s}} \\ & =\frac{33\:\text{m}}{9.45\:\text{s}} \\ & =3.49\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 13

Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice was so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light 3.00×108 m/s.

Solution:

The total distance traveled is computed using the formula

distance=speed×time\text{distance}=\text{speed}\times \text{time}

Therefore, the total distance traveled is

distance=(3.00×108 m/s)(2.56 sec⁡)=768 000 000 meters\begin{align*} \text{distance} & =\left(3.00\times 10^8\:\text{m/s}\right)\left(2.56\:\sec \right) \\ & =768\:000\:000\:\text{meters} \end{align*}

Hence, the distance between the Earth and Moon is equal to the total distance traveled divided by 2.

Earth-Moon Distance=768 000 000 m2=384 000 000 meters=384 000 km  (Answer)\begin{align*} \text{Earth-Moon Distance} & =\frac{768\:000\:000\:\text{m}}{2} \\ & =384\:000\:000\:\text{meters} \\ & = 384 \ 000 \ \text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The distance between the Earth and Moon is approximately 384 thousand kilometers.

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College Physics by Openstax Chapter 2 Problem 12

The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of the axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.1 m long, and the nerve impulse speed is 18 m/s, how long does it take for the nerve signal to travel this distance?

Solution:

The time of travel is computed based on the formula

time=distancespeed\text{time}=\frac{\text{distance}}{\text{speed}}

Therefore, the time of travel is

time=1.1 m18 m/s=0.0611 seconds  (Answer)\begin{align*} \text{time} & =\frac{1.1\:\text{m}}{18\:\text{m/s}} \\ & =0.0611\:\text{seconds} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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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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