Displacement and distance from velocity Consider the velocity ...

Question

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.

d. What is the displacement of the object over the interval [0, 8]?

d. What is the displacement of the object over the interval [0, 8]?

Accepted Answer

Welcome back, everyone. An object moves along a straight path with its velocity function graphed below. The areas between the velocity curve and the t axis over the intervals are indicated. Times in seconds and distance is in meters. Find the displacement of the object over the interval from 0 to 9 inclusive. For this problem, let's recall that displacement is equal to the integral of the velocity function with respect to time between two time values. And in this case we're interested. In time values from 0 to 9. Looking at the graph, we can see that between time of 0 and time of 9, we have 4 segments or 4 different areas. So we can split our integral into 4 parts. The first one is going to be the segment between 0 and 3. The second one is going to be the segment between 3 and 5. Now, let's consider the third one, which is the segment between 5 and 7. And finally we're going to add the integral. From 7 to 9. And now let's recall that definite integrals allow us to identify area between the curve and the horizontal axis. Starting with the first integral, it defines the area between the velocity curve and the t axis. It is equal to 12, and because our area is above the t axis, it is going to be positive. So we're going to add 12 for the first integral. For the second integral, we have area of 18, but our integral is going to have a negative value because our area is below the T axis. So we're going to add -18. Followed by 8 because the segment between 5 and 7. Has area above the T axis. This is where our region is. Meaning we're adding 8th. And finally, our 4th part is going to have a negative sign, that's -4 because our area is below the T axis. So we're going to add -4. Now, let's perform the calculations. We get 12 minus 18, that's -6, 6 plus 8 is 2, and 2 minus 4 is -2. So, we get -2 m. That's our final answer, and thank you for watching.

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