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Chapter 3: Problem 30

A runner times his speed around a track with a circumference of $0.50 \mathrm{mi} .\( He finds that he has run a distance of \)1.00 \mathrm{mi}$ in 4.0 min. What is his (a) average speed and (b) average velocity magnitude in \(\mathrm{m} / \mathrm{s}\) ?

Short Answer

Expert verified
Answer: The runner's (a) average speed is 6.71 m/s, and (b) his average velocity magnitude is 0 m/s.

Step by step solution

01

Convert given units to meters and seconds

We need to convert the given distances from miles to meters and the time from minutes to seconds. We do that by multiplying by the conversion factor. 1 mile = 1609.34 meters 1 minute = 60 seconds Circumference = 0.50 mi * 1609.34 m/mi = 804.67 m Distance traveled = 1.00 mi * 1609.34 m/mi = 1609.34 m Time taken = 4.0 min * 60 s/min = 240 s Now we have all distances in meters and times in seconds.
02

Calculate average speed

Average speed is calculated by dividing the distance run by the time taken. We have the distance traveled (1609.34 m) and the time taken (240 s): Average_speed = (Distance_traveled) / (Time_taken) Average_speed = 1609.34 m / 240 s 鈮� 6.71 m/s The runner's average speed is approximately 6.71 m/s.
03

Calculate the total displacement

Displacement means the change in position from the starting point. Since the runner is moving in a circular track, after covering the 1-mile distance, he has done 2 full laps around the track (1 mile / 0.5-mile circumference). Returning to the starting point after each lap, the total displacement becomes 0. Total_displacement = 0 m
04

Calculate the average velocity magnitude

Since the total displacement is 0, the average velocity magnitude is also 0: Average_velocity_magnitude = (Total_displacement) / (Time_taken) Average_velocity_magnitude = 0 m / 240 s = 0 m/s The runner's average velocity magnitude is 0 m/s. In conclusion, the runner's (a) average speed is 6.71 m/s, and (b) his average velocity magnitude is 0 m/s.

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Most popular questions from this chapter

A speedboat heads west at \(108 \mathrm{km} / \mathrm{h}\) for 20.0 min. It then travels at \(60.0^{\circ}\) south of west at \(90.0 \mathrm{km} / \mathrm{h}\) for 10.0 min. (a) What is the average speed for the trip? (b) What is the average velocity for the trip?

An airplane has a velocity relative to the ground of \(210 \mathrm{m} / \mathrm{s}\) toward the east. The pilot measures his airspeed (the speed of the plane relative to the air) to be \(160 \mathrm{m} / \mathrm{s}\) What is the minimum wind velocity possible?
Prove that the displacement for a trip is equal to the vector sum of the displacements for each leg of the trip. [Hint: Imagine a trip that consists of \(n\) segments. The trip starts at position \(\overrightarrow{\mathbf{r}}_{1},\) proceeds to \(\overrightarrow{\mathbf{r}}_{2},\) then to \(\overrightarrow{\mathbf{r}}_{3}, \ldots\) then to \(\overrightarrow{\mathbf{r}}_{n-1},\) then finally to \(\overrightarrow{\mathbf{r}}_{n} .\) Write an expression for each displacement as the difference of two position vectors and then add them.]
A Nile cruise ship takes \(20.8 \mathrm{h}\) to go upstream from Luxor to Aswan, a distance of \(208 \mathrm{km},\) and \(19.2 \mathrm{h}\) to make the return trip downstream. Assuming the ship's speed relative to the water is the same in both cases, calculate the speed of the current in the Nile.
Demonstrate with a vector diagram that a displacement is the same when measured in two different reference frames that are at rest with respect to each other.
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