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Precalculus : Solve Linear Velocity Problems

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Example Questions

Example Question #1 : Solve Linear Velocity Problems

A plane is traveling Northeast. If the eastern component of it's velocity is $300 \text{ mph}$ , how fast is the plane traveling?

Possible Answers:

$300 \text{ mph}$

None of the other answers

$378 \text{ mph}$

$526 \text{ mph}$

$424 \text{ mph}$

Correct answer:

$424 \text{ mph}$

Explanation:

Since the plane is traveling NE, we know that it is traveling at an angle of 45 degrees north of east. Therefore, we can represent this in a diagram:

We have all of the information we need to solve for the velocity of the plane. We have an angle and a side adjacent to that angle. Therefore, we can use the cosine function to solve for the velocity.

$cos(45^{\circ}) = \frac{300}{v}$

Rearranging for v, we get:

$v = \frac{300}{cos(45^{\circ})} = 424 \text{ mph}$

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Example Question #1 : Angular And Linear Velocity

A car is traveling NW at a speed of mph. What are the magnitudes of each of its North and West components?

Possible Answers:

North: mph

West: 51.96 mph

North: mph

West: mph

North: 84.85 mph

West: 84.85 mph

North: mph

West: mph

North: 42.43 mph

West: 42.43 mph

Correct answer:

North: 42.43 mph

West: 42.43 mph

Explanation:

If the car is traveling NW, it means that the angle $\theta$ is degrees above the x-axis. Using this angle, and the fact that the magnitude of the velocity is mph, we can find the x and y components, or the west and north components using trigonometry.

$\text{West: } \cos(45)=\frac{x}{60}$

x=42.43

$\text{North: } \sin(45)=\frac{y}{60}$

y=42.43

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Example Question #3 : Angular And Linear Velocity

An object is moving in a circular path. How will its linear velocity change if the diameter of the circular path was decreased by one half?

Possible Answers:

$\textup{Decrease by: } \frac{1}{2}$

$\textup{Decrease by: } \frac{1}{4}$

$\textup{Increase by: } \frac{1}{4}$

$\textup{Decrease by: } \frac{1}{8}$

$\textup{Increase by: } \frac{1}{2}$

Correct answer:

$\textup{Decrease by: } \frac{1}{2}$

Explanation:

Write the formula for the linear velocity.

$v=\omega r$

The diameter is twice the radius.

d=2r

If the diameter was decreased by one half, the radius will also decrease by one half.

Therefore:

$v=\omega \cdot \frac{1}{2} r$

$v=\frac{1}{2}( \omega r)$

The linear velocity will decrease by one half.

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Example Question #1 : Solve Linear Velocity Problems

Find the linear velocity in meters per second of an object if it took hours to travel a distance of kilometers. Round to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Write the linear velocity formula.

$v=\frac{x}{t}$

Since the answer is required in meters per second, two conversions will need to be made.

Convert 30 kilometers to meters.

$30 \: \textup{kilometers}\left(\frac{1000\:\textup{meters}}{1 \textup{ kilometer}}\right)=30000 \:\textup{meters}$

Convert 2 hours to seconds.

$2\:\textup{hours}\left(\frac{60\textup{ minutes}}{1\textup{ hour}}\right)\left(\frac{60 \textup{ seconds}}{1 \textup{ minute}}\right)=7200\:\textup{seconds}$

Apply the formula.

$v=\frac{x}{t}=\frac{30000\textup{ meters}}{7200\textup{ seconds}}=4.1\bar{6}=4$

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Example Question #5 : Angular And Linear Velocity

The second hand of a clock is 10.2 cm long. Find the linear speed of the end of this second hand.

Possible Answers:

None of the other answers.

$\frac{3.2cm}{second}$

$\frac{2.1cm}{second}$

$\frac{2cm}{second}$

$\frac{1.07cm}{second}$

Correct answer:

$\frac{1.07cm}{second}$

Explanation:

Linear velocity is defined as distance over a period of time. For instance if a person ran 1 mile or approximately 1600 meters in 7 minutes, the they would have covered about 230 meters per minute. Let's assume this person was running around a track. We could also measure their speed from a central angle and represent their speed as the amount of degrees (or radians) they ran around per unit time as well. This is considered angular speed. A perfect example of both are the hands on a clock. There is a relationship between arc length which we designate s, the radius r, and the angle $\Theta$ (theta). The relationship is $s=r\times\theta$ . So the length of the arc (s) is equal to the radius of the circle the arc is on and what section of the pie it covers ( this is akin to how many degrees our track runner ran "through" or around) $\theta$ . Thus, if we wanted the linear speed around a circle we could say linear speed= s/t or $angular speed=\theta/t$ . Key measurements that you will need to know are how many degrees in a circle of which there are 360 or $2\pi$ . Before you follow the step by step solution below, go back and see if you can use this new information to arrive at the correct answer.

The actual answer is $\frac{1.07cm}{second}$ .

To see why, note that the second hand spins around a total of 360 degrees or $2\pi$ .

$S=20.4cm\times \pi$ .

And how long does it take for the hand to go around? Linear speed of the clock second hand is

$\frac{20.4cm \times \pi}{60s}=$ $\frac{1.07cm}{second}$ (rounded answer).

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Example Question #1 : Solve Linear Velocity Problems

A clock has a second hand measuring 12cm. What is the linear speed of the tip of the second hand?

Possible Answers:

$\frac{2.16cm}{s}$

None of these/

$\frac{6.12cm}{s}$

$\frac{1.62cm}{s}$

$\frac{1.26cm}{s}$

Correct answer:

$\frac{1.26cm}{s}$

Explanation:

Linear speed is equal to the arc length traversed divided by the time. We use s to denote arc length.

$s=2\pi r=2\pi*12cm=24\pi cm$

$LS=\frac{arc length}{t}=\frac{s}{t}= \frac{24\pi cm}{60s}=1.26\frac{cm}{s}$

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Precalculus : Solve Linear Velocity Problems

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Solve Linear Velocity Problems

Example Question #1 : Angular And Linear Velocity

Example Question #3 : Angular And Linear Velocity

Example Question #1 : Solve Linear Velocity Problems

Example Question #5 : Angular And Linear Velocity

Example Question #1 : Solve Linear Velocity Problems

All Precalculus Resources

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