Category Archives: Engineering Mathematics Blog

Problem 1-12: Solving for the percent uncertainty of a 20-m tape that is off by 0.50 cm

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PROBLEM:

A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What is its percent uncertainty?

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SOLUTION:

The percent uncertainty can be calculated by the uncertainty divided by the total measure, but the units should be the same. That is,

% uncertainty=δll×100%=0.5 cm20 m(100 cm1 m)×100%=0.5 cm2000 cm×100%=0.025 %  (Answer)\begin{align*} \%\:\text{uncertainty} & =\frac{\delta _l}{l}\times 100\% \\ \\ & =\frac{0.5\:\text{cm}}{20\:\text{m}\left(\frac{100\:\text{cm}}{1\:\text{m}}\right)}\times 100\% \\ \\ & =\frac{0.5\:\text{cm}}{2000\:\text{cm}}\times 100\% \\ \\ &=0.025\:\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, the percent uncertainty is 0.025%.

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Problem 1-11: Solving for the uncertainty given mass and percent uncertainty

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PROBLEM:

Suppose that your bathroom scale reads your mass as 65 kg with a 3%  uncertainty. What is the uncertainty in your mass (in kilograms)?

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SOLUTION:

The uncertainty is computed as 

δm=3 %100 %×65 kg=1.95 kg  (Answer)\delta _m=\frac{3\:\%}{100\:\%}\times 65\:\text{kg}=1.95\:\text{kg} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Therefore, the uncertainty in the given mass is 1.95 kg.

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Problem 1-10: The average speed of the earth in its orbit in kilometers per second and in meters per second

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PROBLEM:

a) The average distance between the Earth and the Sun is  108 km. Calculate the average speed of the Earth in its orbit in kilometers per second. 

b) What is this is meters per second?

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SOLUTION:

Part A

We assume that the earth revolves around the sun in a circular manner. Therefore, the distance between the earth and the sun will be the radius of its orbit.

The total distance traveled the earth in full revolution is

d=2πr=2π(10)8 km\begin{align*} d & =2\pi r \\ & = 2\pi \left(10\right)^8\:\text{km} \\ \end{align*}

The total time of travel is

t=365.25 days(24 hr1 day)(3600 s1 hr)=3.15576×107 sec\begin{align*} t & =365.25\:\text{days}\left(\frac{24\:\text{hr}}{1\:\text{day}}\right)\left(\frac{3600\:\text{s}}{1\:\text{hr}}\right) \\ & = 3.15576\times 10^7\:\text{sec} \\ \end{align*}

Therefore, the average speed is

speed=distancetime=dt=2π(10)8 km3.15576×107 s=19.91 km/s  (Answer)\begin{align*} \text{speed} & =\frac{\text{distance}}{\text{time}}=\frac{d}{t} \\ & = \frac{2\pi \left(10\right)^8\:\text{km}}{3.15576\times 10^7\:\text{s}} \\ & = 19.91\:\text{km/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, the average speed of the earth is 19.91 km/s.

Part B

Convert 19.91 km/s to m/s.

s=(19.91 kms)(1000 m1 km)=19 910 m/s  (Answer)\begin{align*} s & =\left(19.91\:\frac{\text{km}}{\text{s}}\right)\left(\frac{1000\:\text{m}}{1\:\text{km}}\right) \\ & =19\,910\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Therefore, the velocity is 19 910 m/s. 

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Problem 1-9: The distance moved in 1 sec and the speed in kilometers per million years of tectonic plates

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PROBLEM:

Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year.
(a) What distance does it move in 1.0 s at this speed?
(b) What is its speed in kilometers per million years?

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SOLUTION:

Part A

We know that the formula for the distance traveled, d, if we are given rate, r, and time, t is given by d=r×td=r\times t.

The rate in cm/s is

r=(4 cmyear×1 year365 14 days×1 day24 hours×1 hour60 mins×1 min60 s)=1.2675×10−7 cm/s\begin{align*} r & =\left(4\:\frac{\text{cm}}{\bcancel{\text{year}}}\times \frac{1\:\bcancel{\text{year}}}{365\:\frac{1}{4}\:\bcancel{\text{days}}}\times \frac{1\:\bcancel{\text{day}}}{24\:\bcancel{\text{hours}}}\times \frac{1\:\bcancel{\text{hour}}}{60\:\bcancel{\text{mins}}}\times \frac{1\:\bcancel{\text{min}}}{60\:\text{s}}\right) \\ & = 1.2675 \times 10^{-7} \ \text{cm/s} \end{align*}

The distance traveled is

d=r×td=(1.2675×10−7 cm/s)(1.0 s)d=1.3×10−7 cm  (Answer)\begin{align*} d & = r \times t \\ d & = \left( 1.2675\times 10^{-7} \ \text{cm/s} \right)\left( 1.0 \ \text{s} \right) \\ d & = 1.3\times 10^{-7} \ \text{cm} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We need to convert 4.0 cm/year to km/My.

4.0 cmyear=4.0 cmyear×1 m100 cm×1 km1000 m×1 000 000 years1 My=40 km/My  (Answer)\begin{align*} 4.0 \ \frac{\text{cm}}{\text{year}} & = 4.0\:\frac{\bcancel{\text{cm}}}{\bcancel{\text{year}}}\times \frac{1\:\bcancel{\text{m}}}{100\:\bcancel{\text{cm}}}\times \frac{1\:\text{km}}{1000\:\bcancel{\text{m}}}\times \frac{1\:000\:000\:\bcancel{\text{years}}}{1\:\text{My}} \\ \\ & =40\:\text{km/My} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, the speed of the tectonic plates is 40 km/My. 

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Problem 1-8: The speed of sound in km/h

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PROBLEM:

The speed of sound is measured to be  342 m/s on a certain day. What is this in km/h?

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SOLUTION:

We know that 1 km=1000 m and 1 hr=3600 sec.

Convert 342 m/s.

342 m/s=(342 ms)(1 km1000 m)(3600 s1 hr)=1231.2 km/hr  (Answer)\begin{align*} 342\:\text{m/s} & =\left(342\:\frac{\bcancel{\text{m}}}{\bcancel{\text{s}}}\right)\left(\frac{1\:\text{km}}{1000\:\bcancel{\text{m}}}\right)\left(\frac{3600\:\bcancel{\text{s}}}{1\:\text{hr}}\right) \\ & =1231.2\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, the speed of sound in km/hr is 1231.2 km/hr.

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Problem 1-7: The height of Mount Everest in kilometers

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PROBLEM:

Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its height in kilometers? (Assume that 1 kilometer equals 3,281 feet.)

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SOLUTION:

Convert 29028 ft to km

29,028 ft=(29 028 ft)(1 km3281 ft)=8.847 km  (Answer)29,028\:\text{ft}=\left(29\:028\:\text{ft}\right)\left(\frac{1\:\text{km}}{3281\:\text{ft}}\right)=8.847\:\text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Therefore, the height of Mount Everest is 8.847 kilometers.

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Problem 1-6: The height of a 6 ft 1 in tall person in meters

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PROBLEM:

What is the height in meters of a person who is 6 ft 1.0 in. tall? (Assume that 1 meter equals 39.37 in.)

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SOLUTION:

First, convert 6 ft to inches

6 ft=(6 ft)(12 in1 ft)=72 in6\:\text{ft}=\left(6\:\text{ft}\right)\left(\frac{12\:\text{in}}{1\:\text{ft}}\right)=72\:\text{in}

Add the 1.0 inch

72 in +1 in=73 inches72\:\text{in}\:+1\:\text{in}=73\:\text{inches}

So, the total height of the person is 73 inches. We convert this to meters to come up with the desired unit.

73 in=(73 in)(1 m39.37 in)=1.85 m  (Answer)73\:\text{in}=\left(73\:\text{in}\right)\left(\frac{1\:\text{m}}{39.37\:\text{in}}\right)=1.85\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

So, the height of the person is 1.85 meters.

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Problem 1-5: Soccer field dimensions in feet and inchess

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PROBLEM:

Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its dimensions in feet and inches?

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SOLUTION:

The length in feet and inches are

115 m=115 m×1 ft0.3048 m=377.3 feet  (Answer)115 m=115 m×1 ft0.3048 m×12 inches1 ft=4528 inches  (Answer)\begin{aligned} 115\ \text{m} & = 115\ \bcancel{\text{m}} \times \frac{1\ \text{ft}}{0.3048\ \bcancel{\text{m}}} \\ \\ & =377.3\ \text{feet} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ 115\ \text{m} & = 115\ \bcancel{\text{m}} \times \frac{1\ \bcancel{\text{ft}}}{0.3048\ \bcancel{\text{m}}} \times \frac{12\ \text{inches}}{1\ \bcancel{\text{ft}}} \\ \\ & =4528\ \text{inches} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{aligned}

The width in feet and inches are

85 m=85 m×1 ft0.3048 m=278.9 ft  (Answer)85 m=85 m×1 ft0.3048 m×12 in1 ft=3346 inches  (Answer)\begin{aligned} 85\ \text{m} & = 85\ \bcancel{\text{m}} \times \frac{1\ \text{ft}}{0.3048\ \bcancel{\text{m}}}\\ \\ & =278.9\ \text{ft}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ 85\ \text{m} & = 85\ \bcancel{\text{m}} \times \frac{1\ \bcancel{\text{ft}}}{0.3048\ \bcancel{\text{m}}} \times \frac{12\ \text{in}}{1\ \bcancel{\text{ft}}}\\ \\ & =3346\ \text{inches} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{aligned}
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Problem 1-4: The length of the American football field in meters

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PROBLEM:

American football is played on a 100-yard-long field, excluding the end zones. How long is the field in meters? (Assume that 1 meter equals 3.281 feet.)

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SOLUTION:

100 yard=100 yard×3 feet1 yard×1 m3.281 feet=91.4 m  (Answer)\begin{aligned} 100 \ \text{yard} & = 100 \ \bcancel{\text{yard}} \times \frac{3\ \bcancel{\text{feet}}}{1 \ \bcancel{\text{yard}}}\times \frac{1 \ \text{m}}{3.281\ \bcancel{\text{feet}}} \\ \\ & =91.4 \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{aligned}
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Problem 1-3: Converting 1.0 m/s to 3.6 km/h

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PROBLEM:

Show that 1.0 m/s=3.6 km/h.

Hint: Show the explicit steps involved in converting 1.0 m/s=3.6 km/h.

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SOLUTION:

We know that 1 hr = 3600 s, and 1 km = 1000 m.

1.0 m/s=1.0 ms×3600 s1 hr×1 km1000 m=3.6 km/hr  (Answer)\begin{aligned} 1.0 \ \text{m/s} & = 1.0 \ \frac{\bcancel{\text{m}}}{ \bcancel{\text{s}}} \times \frac{3600 \ \bcancel{\text{s}}}{1 \ \text{hr}}\times \frac{1 \ \text{km}}{1000 \ \bcancel{\text{m}}} \\ \\ & =3.6 \ \text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{aligned}
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