Displacement from velocity The following func...

Question

Displacement from velocity The following functions describe the velocity of a car (in mi/hr) moving along a straight highway for a 3-hr interval. In each case, find the function that gives the displacement of the car over the interval [0,t], where 0 ≤ t ≤ 3.
v(t) = { 30 if 0 ≤ t ≤ 2
50 if 2 < t < 2.5
44 if 2.5 < t ≤ 3

v(t) = { 30 if 0 ≤ t ≤ 2

50 if 2 < t < 2.5

44 if 2.5 < t ≤ 3

Graph showing velocity in mi/hr over time in hours, with distinct segments indicating changes in speed.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says a cyclist's velocity in MPH is described by the following piece-wise function for a 2-hour ride. Find the function that gives the displacement of the cyclists over the close interval from 0 to T, where 0 is less than equal to T, which is less than or equal to 2. And we're given. VFT is equal to 18, if 0 is less than or equal to T, which is less than or equal to 1, 24 if 1 is less than T, which is less than or equal to 1.5, and 20 if 1.5 is less than T, which is less than or equal to 2. What is the displacement function as of T? Now, to begin with, we need to call that our displacement functions of T is going to be equal to the integral from 0 to T of VX. DX. So our displacement here is just our interval of our velocity function. However, since our velocity function is a piecewise function, we're going to have to look at 3 separate cases. So, the first case that we're going to look at is when 0 is less than or equal to T, which is less than or equal to 1. So, for this case, we will have that S of T is going to be equal to the integral from 0 to T. Of, and here we just need to use the first piece of our piecewise function for VFX since T is between 01. So, we'll have 18. DX for the integra. So now we just need to perform the integration. So this is going to be equal to, and here we call that when you integrate 18 with respect to X, you get 18 X. This needs to be evaluated from 0 to T. Substituting in our limits of integration, this is going to be equal to 18, multiplied by the quantity of T minus 0 in quantity and simplifying, we get that SF T is equal to 18 T. Now, for the second case, we need to look at when 1 is less than T, which is less than or equal to 1.5. And for this case, SFT is going to need to be broken up into two integrals. So SFT is going to be equal to the integral from 0 to 1 of the first piece of our piece-wise function, which is going to be 18, and that gets multiplied by DX. Then we'll have plus the integral from 1 to T of the second piece of our piecewise function, which is going to be 24, and that gets multiplied by DX. So performing the integration, this is going to be equal to. And as before we integrate 18 with respect to X, we get 18 X. This needs to be evaluated from 0 to 1. Then we'll have plus, and when we integrate 24 with respect to X, we get 24 X, and this needs to be evaluated from 1 to T. And so substituting in our limits of integration, this is going to be equal to 18, multiplied by the quantity of 1 minus 0 in quantity, plus 24, multiplied by the quantity of T minus 1 in quantity. And so by this expression, we see that this is going to be equal to 24 T minus 6. Now we can factor a 6 out of both of those terms. In doing so, we see that SOT is going to be equal to 6 multiplied by the quantity of 4 T minus 1 in quantity. And the final case that we need to look at is when 1.5. is less than T, which is less than or equal to 2. So, in this case, we will again need to break up SFT, but we'll have to break it up into three integrals, one for each piece of our piece-wise function. So S of T is going to be equal to the integral from 0 to 1 of 18 DX plus the integral from 1 to 1.5 of 24 DX. Then we'll have plus the integral from 1.5 to T. Of the last piece of our piece wise function, which was 20, and that gets multiplied by DX. So, again, we need to perform the integration. So this is going to be equal to. Again, we integrate 18 with respect to X, we get 18 X. This needs to be evaluated from 0 to 1. Then we'll have plus, and again, when we integrate 24 with respect to X, we get 24 X, and this needs to be evaluated from 1 to 1.5. And then finally, we'll have plus. When we integrate 20 with respect to X, we will have 20 X, and this needs to be evaluated from 1.5 to T. So substituting in the limits of integration, we see that this is going to be equal to 18, multiplied by the quantity of 1 minus 0 in quantity, plus 24 multiplied by the quantity of 1.5 minus 1 in quantity, then plus 20 multiplied by the quantity of T minus 1.5 in quantity. And so finding this expression, we see that this is going to be equal to 20T. So, for our final answer, we will have that SFT. is going to be equal to, and here again, we'll have a piecewise function. So, for the first piece, we will have 18T. If 0 is less than or equal to T, which is less than or equal to 1, we will have 6 multiplied by the quantity of 4 T minus 1 in quantity. If 1 is less than T, which is less than or equal to 1.5. And 2 tea. If 1.5 is less than T, which is less than or equal to 2, so this here is the answer that we were looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Displacement as the Integral of Velocity

Piecewise Functions and Integration

Interpreting Graphs of Piecewise Velocity Functions

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