Displacement and distance from velocity Consider the velocity function shown below of an object moving along a line. Assume time is measured in seconds and distance is measured in meters. The areas of four regions bounded by the velocity curve and the t-axis are also given.

b. What is the displacement of the object over the interval [2, 6]? 

Verified step by step guidance

1. Identify the interval over which you need to find the displacement, which is from \(t=2\) to \(t=6\) seconds.
2. Recall that displacement over a time interval is the net change in position, which can be found by integrating the velocity function over that interval: \(\text{Displacement} = \int_{2}^{6} v(t) \, dt\).
3. Use the given areas under the velocity curve between \(t=2\) and \(t=6\). From the graph, the velocity is positive from \(t=2\) to \(t=4\) with an area of 14, and negative from \(t=4\) to \(t=6\) with an area of 10.
4. Since the velocity is positive on \([2,4]\), the displacement contribution is +14 meters, and since the velocity is negative on \([4,6]\), the displacement contribution is -10 meters.
5. Add these signed areas to find the total displacement over \([2,6]\): \(14 + (-10) = 14 - 10\) meters.

Key Concepts

Velocity and Displacement

Velocity is the rate of change of position with respect to time and can be positive or negative depending on direction. Displacement over a time interval is the net change in position, calculated as the integral of velocity over that interval, which corresponds to the signed area under the velocity curve.

Definite Integral as Area Under a Curve

The definite integral of a function over an interval represents the net area between the function's graph and the horizontal axis. For velocity, positive areas indicate movement in one direction, while negative areas indicate movement in the opposite direction, affecting the total displacement.

Interpreting Signed Areas for Displacement

When calculating displacement from velocity, areas above the time-axis contribute positively, and areas below contribute negatively. Summing these signed areas over the interval gives the object's net displacement, reflecting both magnitude and direction of movement.