Question

1. a cyclist is initially traveling along a level, straight road at 10 m/s in the positive direction. the magnitude of the velocity for the cyclist is tracked from this point forward using the graph. during which time interval did the cyclist have the greatest acceleration? clear all 0 - 1 seconds 1 - 2 seconds 2 - 3 seconds 3 - 4 seconds velocity - time graph velocity (m/s) 0 5 10 15 20 1 2 3 4 5 time (s)

Explanation:

Step1: Recall acceleration - velocity - time relation

Acceleration $a=\frac{\Delta v}{\Delta t}$, and on a velocity - time graph, acceleration is the slope of the graph.

Step2: Calculate slopes for each interval

For 0 - 1 seconds, velocity is constant, so $a_1 = 0$. For 1 - 2 seconds, $\Delta v_2=0$, so $a_2 = 0$. For 2 - 3 seconds, if we assume initial velocity $v_1$ at $t = 2s$ and final velocity $v_2$ at $t=3s$, from the graph, $\Delta v_2$ and $\Delta t_2 = 1s$. For 3 - 4 seconds, the slope is steeper than in 2 - 3 seconds. Let the initial velocity at $t = 3s$ be $v_{i}$ and final velocity at $t = 4s$ be $v_{f}$. The change in velocity $\Delta v=\vert v_{f}-v_{i}\vert$ in 3 - 4 seconds is larger for the same time - interval $\Delta t = 1s$ compared to other non - zero slope intervals.

Answer:

3 - 4 seconds