9–10. Velocity graphs The figures show velocity functions for motion along a line. Assume the motion begins with an initial position of s(0)=0. Determine the following.

a. The displacement between t=0 and t=5

Verified step by step guidance

1. Identify that displacement is the integral of velocity over time, which geometrically corresponds to the area under the velocity-time graph between the given time interval.
2. Divide the graph into simple geometric shapes between t = 0 and t = 5. Here, the graph consists of a rectangle from t = 0 to t = 3 with height 3, and a triangle from t = 3 to t = 5 where velocity decreases linearly from 3 to 0.
3. Calculate the area of the rectangle: Area = base \(\times\) height = (3 - 0) \(\times\) 3.
4. Calculate the area of the triangle: Area = \(\frac{1}{2}\) \(\times\) base \(\times\) height = \(\frac{1}{2}\) \(\times\) (5 - 3) \(\times\) 3.
5. Add the areas of the rectangle and triangle to find the total displacement between t = 0 and t = 5.

Key Concepts

Velocity and Displacement

Velocity represents the rate of change of position with respect to time. Displacement is the net change in position over a time interval and can be found by integrating the velocity function over that interval. Positive velocity indicates movement in the positive direction, while negative velocity indicates movement in the opposite direction.

Area Under the Velocity-Time Graph

The displacement over a time interval is equal to the area under the velocity-time graph between those times. For piecewise linear velocity graphs, this area can be calculated by summing the areas of geometric shapes like rectangles and triangles formed under the curve.

Piecewise Functions and Integration

When velocity changes in segments, the function is piecewise. To find displacement, integrate each segment separately and sum the results. This approach simplifies calculations, especially when velocity is constant or changes linearly within intervals.