Question

activity c: distance and displacement
get the gizmo ready

• turn off show graph and show animation for runner 2

1. create the position - time graph for runner 1 shown at right. then fill in the blanks below to describe what you think the runner will do, based on that graph.

the runner will run __ yards in the first 2 seconds, with a velocity of __ y/s
his direction will be from __ to __
then he will run __ yards in the next 2 seconds, with a velocity of __ y/s.
his direction will be from __ to __
click the green start button and watch the runner go. were you correct? ____

1. two students, gina and walter, are discussing the runner whose graph is shown above.

• gina says the runner moved more than 40 yards
• walter says the runner moved less than 40 yards.

a. who do you think is right? ____
b. explain your answer. ____

1. on top of the left half of the gizmo, select the distance traveled tab.

a. what was the total distance traveled by the runner after 4 seconds? ____
b. displacement is equal to the difference between the starting and ending positions. displacement to the right is positive while displacement to the left is negative.
what is the displacement shown by the graph at the top of the page? ____
(activity c continued on next page)

Explanation:

Step1: Analyze first 2 - second interval

On a position - time graph, the distance traveled is the change in position. If the runner starts at position 0 and at \(t = 2\) seconds is at position 20 yards (assuming from the graph), the distance \(d_1=20\) yards. The velocity \(v_1=\frac{\Delta x}{\Delta t}\), where \(\Delta x = 20\) yards and \(\Delta t=2\) seconds, so \(v_1 = 10\) y/s. The direction is from the starting point (assume left - most position on the graph) to the right.

Step2: Analyze second 2 - second interval

If the runner moves from 20 yards to 30 yards in the next 2 seconds (\(t = 2\) to \(t = 4\) seconds), the distance \(d_2=10\) yards. The velocity \(v_2=\frac{\Delta x}{\Delta t}=\frac{30 - 20}{2}=5\) y/s. The direction is from the 20 - yard position to the 30 - yard position (still to the right).

Step3: Answer part 2A

The total distance traveled by the runner is \(d=d_1 + d_2=20+10 = 30\) yards. So Walter is right.

Step4: Answer part 2B

The total distance is the sum of the distances in each interval. In the first 2 seconds, the runner moves 20 yards and in the next 2 seconds, the runner moves 10 yards, for a total of 30 yards which is less than 40 yards.

Step5: Answer part 3A

The total distance traveled after 4 seconds is 30 yards as calculated above.

Step6: Answer part 3B

If the starting position is 0 yards and the ending position is 30 yards, the displacement \(\Delta x=30 - 0=30\) yards.

Answer:

1. The runner will run 20 yards in the first 2 seconds, with a velocity of 10 y/s. His direction will be from left to right. Then he will run 10 yards in the next 2 seconds, with a velocity of 5 y/s. His direction will be from 20 - yard position to 30 - yard position.
2. A. Walter

B. The total distance is the sum of the distances in each interval. In the first 2 seconds, the runner moves 20 yards and in the next 2 seconds, the runner moves 10 yards, for a total of 30 yards which is less than 40 yards.

1. A. 30 yards

B. 30 yards