Category Archives: Physics

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\:\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\:\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\:\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

\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

\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:

\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. 

\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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Problem 1-2: Converting car speed of 33 m/s to kilometers per hour and determining if it exceeds the speed limit

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PROBLEM:

A car is traveling at a speed of 33 m/s.
(a) What is its speed in kilometers per hour?
(b) Is it exceeding the 90 km/h speed limit?

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SOLUTION:

Part A

\begin{aligned}
33 \ \text{m/s} & =33\ \frac{\text{m}}{\text{s}} \times \frac{1\ \text{km}}{1000 \ \text{m}} \times \frac{3600\ \text{s}}{1 \ \text{hr}} \\
\\
& =118.8 \ \text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{aligned}

Part B

At 118.8 km/h, the car is traveling faster than the speed limit of 90 km/h. (Answer)

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Problem 1-1: Converting 100 km/h to meters per second and miles per hour

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PROBLEM:

The speed limit on some interstate highways is roughly 100 km/h.
(a) What is this in meters per second?
(b) How many miles per hour is this?

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SOLUTION:

Part A

\begin{aligned}
100 \  \frac{ \text{km}}{\text{hour}} & =100 \ \frac{ \bcancel{\text{km}}}{\bcancel{\text{hour}}} \times \frac{1000 \ \text{m}}{1 \ \bcancel{\text{km}}} \times \frac{1 \ \bcancel{\text{hour}}}{3600 \ \text{sec}}\\
\\
&=27.7 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{aligned}

Part B

\begin{aligned}
100 \  \frac{ \text{km}}{\text{hour}} & =100 \ \frac{ \bcancel{\text{km}}}{\text{hour}} \times\frac{1 \ \text{mile}}{1.609\ \bcancel{\text{km}}} \\
\\
&=62.2 \ \text{mi/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{aligned}
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Video Solution:

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