Question

Below you will see diagrams showing a trip taken by Fred and a trip taken by Frank. For each person, find their distance, displacement, average speed, and average velocity from A to B. Each square is 10 meters wide, and use words like north, south, east, and west for direction.

The first thing Fred did was go south for 4 seconds. If he wanted to cut his average speed in half, how long should that southward motion last (assuming the other two motions happen in the same time as shown)? Explain your reasoning.

Would this amount of wait-time also have halved his average velocity?

Answer 1

Part A

The completed table of values of the motion of Fred and Frank are;

$\begin{array}{|l|c|c|}&\mathbf{Fred's \ Trip}&\mathbf{Frank's \ Trip}\\\mathbf{Distance}&150 \, m&120 \, m\\\mathbf{Displacement}&30 \, m&50 \, m\\\mathbf{Average \ Speed}&15 \, m/s&10 \, m/s\\\mathbf{Average \ Velocity}&3 \ m/s&4.1 \overline 6 \, m/s\end{array}\right]$

Part B

The time the southward motion should last is 14 seconds

The reason the above values are correct is as follows;

The total distance is given by summing the length of the sides of the square that lay along the path

Known parameters:

For Fred's Trip, we have;

Distance = 60 meters South + 30 meters East  + 60 meters north = 150 meters The total distance = 150 meters

Required:

The displacement = 30 meters East

Average speed = Total distance/(Total time)

Total time = 4 sec + 3 sec + 3 sec = 10 sec

Average speed = 150 m/10 sec = 15 m/s

Average velocity = Displacement/Total time

Average velocity = 30 m/10 sec = 3 m/s

Known parameters:

For Frank's Trip, we have;

Distance = 30 meters North + 60 meters East  + 30 meters south = 120 meters

The total distance of Frank's trip = 120 meters

Required:

The displacement = 50 meters East

Average speed = Total distance/(Total time)

Total time = 5 sec + 5 sec + 2 sec = 12 sec

Average speed = 120 m/12 sec = 10 m/s

Average velocity = Displacement/Total time

Average velocity = 50 m/12 sec = 4.1$\overline 6$ m/s

Therefore, we have;

$\begin{array}{|l|c|c|}&\mathbf{Fred's \ Trip}&\mathbf{Frank's \ Trip}\\\mathbf{Distance}&150 \, m&120 \, m\\\mathbf{Displacement}&30 \, m&50 \, m\\\mathbf{Average \ Speed}&15 \, m/s&10 \, m/s\\\mathbf{Average \ Velocity}&3 \ m/s&4.1 \overline 6 \, m/s\end{array}\right]$

To reduce his speed in half, we have;

New average speed, s₂ = s₁/2

∴ s₂ = (15 m/s)/2 = 7.5 m/s

Let x represent the time for the southward motion, we have;

(150)/(6 + x) = 7.5

(6 + x) × 7.5 = 150

x = (150 - 7.5×6)/7.5 = 14

Therefore, the time for the southward motion should last for 14 seconds

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