Question

distance and displacement during one play in the soccer game on the previous page, the player runs 41.2 m from position d to position c. then she runs 10 m to position b. her path is shown by the green dotted lines. the total distance the player travels is 41.2 m + 10 m = 51.2 m. the solid green arrow in the figure shows the player’s displacement. displacement is the difference between the initial, or starting, position and the final position. the player starts at point d and finishes at point b. her displacement is 40 m in front of her initial position. displacement is the shortest distance between where the player started and the player’s final position. an object’s displacement and the distance it travels are not always equal. if the player runs directly from point d to point a, then both the player’s distance and displacement are the same—10 m. if the player’s final position is the same as her starting position, her displacement is 0 m. three - dimensional thinking use the race track model below to determine the distance traveled and the displacement of a car from point a to when it reached point d on the first lap.

Explanation:

Step1: Identify point D's position

Point D is the end of the left curved section, opposite the 100 m segment. First, calculate the length of the curved (semicircular) part of the track. The straight side length is 220 m, so the diameter of the semicircular ends is $580\ \text{m} - 220\ \text{m} = 360\ \text{m}$. The length of one semicircular end is $\frac{1}{2} \times \pi \times 360\ \text{m} = 180\pi\ \text{m} \approx 565.49\ \text{m}$.

Step2: Calculate total distance traveled

From A to D: travel 100 m to the end of the straight segment, plus the full length of the opposite straight segment (220 m), plus the left semicircular end (≈565.49 m).
$\text{Distance} = 100\ \text{m} + 220\ \text{m} + 565.49\ \text{m} = 885.49\ \text{m}$

Step3: Calculate displacement

Displacement is the straight-line distance from A to D. A is 100 m from the right end, so the horizontal distance from A to the center line is $100\ \text{m}$. The vertical distance from A to D is the full width of the track, 360 m. Use the Pythagorean theorem:
$\text{Displacement} = \sqrt{(100\ \text{m})^2 + (360\ \text{m})^2} = \sqrt{10000 + 129600} = \sqrt{139600} \approx 373.63\ \text{m}$

Answer:

Distance traveled: $\approx 885.5\ \text{m}$
Displacement: $\approx 373.6\ \text{m}$