Answer:
a. The total distance traveled during the five seconds is the sum of the lengths of all the line segments on the graph:
Distance traveled = AB + BC + CD + DE = 3 m + 4 m + 2 m + 5 m = 14 m
b. The displacement between t=1s and t=5s is the straight-line distance between the starting point (A) and the ending point (E):
Displacement = AE = 5 m - 0 m = 5 m (to the right)
c. The average velocity from t=0s to t=3s is the ratio of the displacement to the time interval:
Average velocity = displacement / time interval = DE / 3 s = 5 m / 3 s ≈ 1.67 m/s (to the right)
d. The greatest speed occurs during the line segment with the steepest slope, which is BC.
a. The distance traveled during all 5 seconds is the sum of the lengths of all the line segments on the graph:
Distance traveled = AB + BC + CD + DE + EF = 3 m + 1.5 m + 1.5 m + 2 m + 2 m = 10 m
b. The average velocity during the full five seconds is the total displacement divided by the time interval:
Average velocity = displacement / time interval = 0 m / 5 s = 0 m/s
(Note: The displacement is zero because the boy ends up at the same position as where he started.)
c. To find the instantaneous speed at t=1s, we need to find the slope of the tangent line to the position-time graph at t=1s. From the graph, we see that the position changes by 2 m during a time interval of 1 second, so the instantaneous speed at t=1s is:
Instantaneous speed at t=1s = 2 m/s