93. Three start-ups Three cars, A, B, and C, start from rest a...

Question

93. Three start-ups Three cars, A, B, and C, start from rest and accelerate along a line according to the following velocity functions:

vₐ(t) = 88t/(t + 1), v_B(t) = 88t²/(t + 1)², and v_C(t) = 88t²/(t² + 1).

b. Which car travels farthest on the interval 0 ≤ t ≤ 5?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says two trains, P and Q, start from rest and accelerate along a line according to the following velocity functions in meters per second. And we're given that visa PFT is equal to 75T divided by the quantity of T + 3 in quantity, and visaq of T is equal to 75T 2. Divided by the quantity of T2 + 9 in quantity, where T is the time in seconds. Which train travels farther on the interval? 0 is less than or equal to T, which is less than equal to 4. And we're given two possible choices as our answers. For choice A, we have train P, and for choice B, we have train Q. Now, in order to see which train travels farther on our time interval, we just need to calculate the displacement of each train. So we're gonna start by looking at train P, and we're gonna call its displacement X sub P. And we called that this plate is going to be equal to the interval over a time interval, which is going to go from 0 to 4, of visa PFT, which is 75 T, divided by the quantity of T + 3 in quantity, and we need to multiply this by DT since we're going to be integrating with respect to time. Now, in order to perform this integration, we want to break our fraction up here into two separate terms, using partial fractions. And so, we call for partial fractions that 75 T divided by the quantity of T + 3 in quantity, it's going to be equal to A plus B divided by the quantity of T + 3 in quantity. So the next step is to multiply both sides of the equation by the quantity of T + 3 in quantity, and then solve for A and B. So performing the multiplication, we'll have that 75 T is equal to a multiplied by the quantity of T + 3 and quantity plus B. Multiplying everything out on the right hand side. We'll have that 75 T is equal to AT plus 3A plus B. So now we just need to match the coefficients for the various powers of T. So for the linear term, we'll have that A is equal to 75, and for the constant term, we'll have that 3A plus B is equal to 0, since we have no constant term on the left-hand side of our equation. So we can take the value for A from the first equation, which is A is equal to 75, and substitute it into the second equation to solve for B. So doing so, we'll have that 3 multiplied by 75, plus B is equal to 0, and solving for B, we get that B is equal to minus 225. So that means that we can rewrite our displacement X subp as being equal to the integral from 0 to 4 of 75 DT plus the interval from 0 to 4 of minus 225 divided by the quantity of T + 3 and quantity dT. So now we just need to perform the integration. So that means that our displacement here is going to be equal to, and here for the first integral, we're going to be integrating 75 with respect to T. recall that that's going to give us 75 T and this needs to be evaluated from 0 to 4. And then we'll have minus 225, and this gets multiplied by, and here we're going to be integrating 1 divided by the quantity of T + 3 in quantity with respect to T. recall that's going to give us the natural log of the absolute value of the quantity of T + 3 in quantity, and this again needs to be evaluated from 0 to 4. So substituting in our limits of integration, this is going to be equal to 75 multiplied by 4. Minus 75, multiplied by 0. Minus 225, multiplied by the quantity of the natural log of the absolute value of 4 + 3. And then we'll have minus the natural log of the absolute value of the quantity of 0 plus 3 in quantity. So simplifying this expression, this is going to be equal to. 300 Minus 225 multiplied by the natural log of the quantity of 7 divided by 3 in quantity. So here for the last term we've combined our two natural log functions using in properties of logarithms, and we've also dropped the absolute value signs since both 7 and 3 are positive quantities. And so when we evaluate this expression in our calculator, we see that our displacement is going to be approximately equal to. 109.4 m. So here we just round it to one decimal place. So we're gonna have to follow a similar procedure in order to find the displacement for train Q. So we'll call that displacement X subq, and so xq is going to be equal to the integral from 0 to 4 of V subq, which is 75 T 2 divided by the quantity of T2 + 9 in quantity, and this gets multiplied by dT. This is going to be again integrating with respect to time. And like with the first fraction, we want to separate this fraction up into multiple terms using partial fractions. So again, using partial fractions we'll have that 75 t squared divided by the quantity of T2 plus 9, and quantity is going to be equal to a plus B divided by the quantity of T2 plus 9 in quantity. And again, we need to solve for A and B. So multiplying both sides of the equation by the quantity of T2 + 9 in quantity, we'll get that 75 T2 is equal to a. Multiplied by the quantity of T2 + 9 in quantity plus B. Multiplying everything out on the right hand side, we'll have that 75 T squared is equal to A T2 + 9 A plus B. Again, matching up the coefficients for the various powers of T for the quadratic term, we'll have that A is equal to 75, and for the constant term we'll have that 9 A plus B is equal to 0. So again, substituting the value for for A, which is equal to 75, into the second equation, and solving for B will have that 9 multiplied by 75, plus B is equal to 0, so that means that B is going to be equal to minus 675. So that means that we can rewrite our displacement integral. So x subq is going to be equal to the integral from 0 to 4 of 75 dt plus the integral from 0 to 4 of minus 675 divided by the quantity of T2 plus 9 in quantity dt. So, performing the integration, this is going to be equal to, for the first interval, we're going to be integrating 75 with respect to T again, and so that's going to again give us 75 T. This needs to be evaluated from 0 to 4. They'll have minus 675, multiplied by, and here we're going to be integrating 1 divided by the quantity of T2 + 9 in quantity with respect to T. And we call that when we perform that integration, we get the quantity of 1 divided by 3 multiplied by the inverse tangent of the quantity of T divided by 3 in quantity, and this needs to be evaluated from 0 to 4 as well. So substituting in our limits of integration, we see that our displacement X subq is going to be equal to 75 multiplied by 4. Minus 75, multiplied by 0. Minus and here we can simplify the coefficient, so we'll have 675 divided by 3, which is 225. And then here we're going to be multiplying this by the quantity of the inverse tangent of the quantity of 4 divided by 3 in quantity minus the inverse tangent of the quantity of 0 divided by 3 in quantity. So simplifying this expression, this is going to be equal to. 300 Minus 225 multiplied by the inverse tangent of the quantity of 4 divided by 3 in quantity and when we evaluate this expression using our calculator, we see that our displacement is going to be approximately equal to. 91.4 m. So now we just need to determine which one of these is the larger value. And so if we look at our two values, we see that the larger value corresponds to train P. So that means that the answer for this problem is going to be that train P travels the further distance. So that is the answer that we're looking for for this problem. And when we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

93. Three start-ups Three cars, A, B, and C, start from rest and accelerate along a line according to the following velocity functions:
vₐ(t) = 88t/(t + 1), v_B(t) = 88t²/(t + 1)², and v_C(t) = 88t²/(t² + 1).
b. Which car travels farthest on the interval 0 ≤ t ≤ 5?

Key Concepts

Velocity and Displacement Relationship

Definite Integration

Analyzing Rational Functions

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