Displacement from a velocity graph Consider t...

Question

Displacement from a velocity graph Consider the velocity function for an object moving along a line (see figure).

(c) Use geometry to find the displacement of the object between t = 2 and t = 5.

Velocity graph showing time in seconds on the x-axis and velocity in meters per second on the y-axis.

Accepted Answer

Welcome back, everyone. Given the velocity time graph of a car moving along a straight road for a T in the interval from 0 to 10, what is the displacement of the particle between T equals 5 and the T equals 8? Here we have the velocity time graph and for our answer choices A says it's 30 m, B 50 m, C 20 m, and the D 80 m. Now how can we find the displacement of our particle? Well, recall that the displacement is going to be the area under the velocity time graph from the times that we specify. In this case, between T equals 5 and T equals 8. So if we go to our graph and go to where T equals 5, all the way up to where T equals 8, then we're trying to find the area of the curve in this section. And then let me put some vertical lines above our times here for us to better see exactly what we're looking for. Now, if you notice here that for this area, it's basically split into two parts on this interval, right? Now, if I were to draw a line here where T equals 6, then you notice that, OK, let me make this a dotted line, you'll notice that it's made up of. A rectangle between T equals 5 to T equals 6. OK, that's right here. And then it's made up of a trapezoid between T equals 6 to T equals 8. OK, that I can basically just draw right here. So essentially what we're doing then we need to find the area for both of these ships to help us find the displacement. First, let's start with the area from T equals 5. The T equals 6. No, that means we're finding the area of a rectangle, and to find the area of a rectangle we multiply its length by its width. And for a velocity time graph, its length, or sorry, its height, well, you know what, to be clear, let's say the base by the height where our base is our time and our height is our velocity. So in this case, the displacement. It is going to be equal to 30 m per second. Its height multiplied by 1 2nd, the length of its base, because from 5 to the 5th to the 6th second is 1 2nd. And when we do that, then we get the displacement to be 30 m. Now, we can try to figure out the displacement from T equals 6 to T equals 8. That means we're finding the area of a trapezoid. And to find the area of a trapezoid, first, it would be a good idea for us to make note of each point. So, let's call the point where T equals 6, T1, V1, OK, where T1 is 6 and V1 is 30. Well, the second point we can call T2V2, where T2 is 8 and V2 is 20. No, for the era of a chopper side, OK. The displacement. The area under the curve will be equal to a half of the sum of V1 and V2, OK, divided by T2 minus T1. In that case, that's going to be a half of 30 plus 20. OK, 30 + 20 m per second, divided by 8 minus 6 seconds. Now, 30 + 20 is 50, OK? So that's 50 m per 2nd and 8 minus 6 is 2. So, when we go ahead and multiply here, we get our displacement to be 50 m. Thus, from T equals 6 to T equals 8, the displacement is 50 m. Therefore, That means then that our displacement. From T equals 5 to T equals 8 is going to be equal to 30 m plus 50 m, which equals 80 m. Thus, that is going to be the displacement of our particle between the given times. If we look back at our answer twices, that means D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Displacement from a velocity graph Consider the velocity function for an object moving along a line (see figure).
(c) Use geometry to find the displacement of the object between t = 2 and t = 5.

Key Concepts

Velocity Function

Displacement

Area Under the Curve

Watch next