Example 16  A Bicycle Trip

A bicyclist maintains a constant velocity on the outgoing leg of a trip, zero velocity while stopped, and another constant velocity on the way back. Figure 2.22 shows the corresponding position-time graph. Using the time and position intervals indicated in the drawing, obtain the velocities for each segment of the trip.

This position-vs.-time graph consists of three straight-line segments, each corresponding to a different constant velocity.
Figure 2.22  This position-vs.-time graph consists of three straight-line segments, each corresponding to a different constant velocity.

Reasoning  The average velocity is equal to the displacement Dx divided by the elapsed time . The displacement is the final position minus the initial position, which is a positive number for segment 1 and a negative number for segment 3. Note for segment 2 that Dx=0 m, since the bicycle is at rest. The drawing shows values for Dx and Dt for each of the three segments.

Solution The average velocities for the three segments are

In the second segment of the journey the velocity is zero, reflecting the fact that the bike is stationary. Since the position of the bike does not change, segment 2 is a horizontal line that has a zero slope. In the third part of the motion the velocity is negative, because the position of the bike decreases from x=+800 m to x=+400 m during the 400-s interval shown in the graph. As a result, segment 3 has a negative slope, and the velocity is negative.



Copyright © 2000-2003 by John Wiley & Sons, Inc. or related companies. All rights reserved.