A slowing race Starting at the same time and place, Abe and Bo...

Question

A slowing race Starting at the same time and place, Abe and Bob race, running at velocities u(t) = 4 / (t + 1) mi/hr and v(t) = 4e^(−t/2) mi/hr, respectively, for t ≥ 0.

b. Find and graph the position functions of both runners. Which runner can run only a finite distance in an unlimited amount of time?

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Emily and Frank start a walk at the same time from the same place. Emily's velocity is U of T is equal to 5 divided by T + 3 m per second and Frank's velocity is V of T is equal to 5 multiplied by E to the power of negative T divided by 5 m per second, or T is greater than or equal to 0. Find the position functions for both walkers. Which walker can travel only a finite distance as T approaches positive infinity. Take 0 as the initial position. Awesome. So it appears for this particular problem we're asked to solve for 3 separate answers. Our first two answers that we are asked to solve for is we're asked to determine what Emily's position function is, and our second answer is we're asked to solve for Frank's position function based on the information that's provided to us by the problem itself. And lastly, our final answer we're asked to solve for is which walker is it, Emily or Frank, can travel only a finite distance as T approaches positive infinity. So will it be Emily or Frank who can Travel only a finite distance. So with that in mind, now that we know what we're ultimately trying to solve for, our first step is we're gonna have to assume that both start from position or start at a position of 0. At T equals 0 or T denotes the time. And they're gonna start as start at this position, which I'll put 0 in quotation marks, a position of zero since they start from the initial position will be 0, which means start from rest, they're not moving at all, and then they start to move. So, with that in mind, we need to note the following. We need to recall and note the following. So the position S of T, so our position function will be S of T. So the positions of T. With a S of 0 will be equal to 0. So when T is equal to 0, S is 0 will be equal to 0. Is that S of T will be equal to the integral from 0 to T of W X. DX for the velocity W. Fantastic. So now focusing, so this would be our first step. Our second step now is we're going to be focusing on Emily first, so focusing our attention strictly on Emily, and we're going to deal with Frank's information in a minute, but focusing on Emily, we will note that the velocity U of T is equal to 5 divided by T + 3. So we need to compute that S subscript E, which I'll call SE of T. So S E T, so this is the position function for Emily. So S E of T is equal to the integral from 0 to T of 5 divided by X + 3. DX and now we need to factor out the constant. So when we do, we will find that SE of T is equal to 5 multiplied by the integral from 0 to T1 divided by X plus 3. DX. So to keep things simple, I just turn the T into an X so we can do the integral or solving for the integral a little bit more simple like. So with that in mind, Moving right along here, our next step now after we factored the constant is we need to recall and use the substitution method. So we need to substitute, and let's make a note, we need to substitute. Y is equal to x + 3. So that will mean that D Y is equal to DX and the limits as X is equal to 0. We can state that Y will be equal to 0, and as T is X is equal to T, so as X is equal to T, we can say that Y. is equal to T + 3. So with that in mind, we can go ahead and safely write that SE of T is equal to 5 multiplied by the integral from 32 T plus 3 of 1 divided by Y D Y, which will be equal to 5 multiplied by square brackets of the natural log, the absolute value of Y, evaluated from 3 to T + 3. Fantastic. So thus we can go ahead and write that SE. Of T, so SE of T will be equal to 5 multiplied by parenthesis, the natural log of T + 3 minus the natural log of 3. Which will be equal to 5 multiplied by the natural log of T + 3 divided by 3. And of course, since we're dealing with position and we're told that the velocities in units and meters per second, we will note that our units for this particular case will be M meters. Fantastic, and that is our first answer. Hooray, we did it. So now we can start to solve for Frank. So now, moving on to step 3, solving for Frank. So let's make a note that we're solving for Frank now. We need to note and recall that we're told that VAT, so Frank's velocity, is equal to 5 multiplied by E to the power of negative T divided by 5, so we need to compute that. SF of T is equal to the integral from 0 to T of 5 multiplied by E to the power of negative X divided by 5 D X. So now we need to factor out the constant. So SF of T will be equal to 5 multiplied by the integral of 0 to T of E to the power of negative X divided by 5DX. And at this stage, we need to recall that the antiderivative of E to the power of negative X divided by 5 is -5 multiplied by E to the power of negative X divided by 5. Sense D by DX parentheses negative 5 multiplied by e to the negative x divided by 5 will be equal toe the power of negative x divided by 5. Fantastic. So with that knowledge in mind, we can therefore write that SF of T. Will be equal to 5 multiplied by square brackets of negative 5 multiplied by e to the power of negative x divided by 5 evaluated from 0 to T, which will be equal to 5 multiplied by parentheses -5 multiplied by e to the power of negative T divided by 5 plus 5. Multiplied by e to the power of 0. So when we simplify a little bit, we will find that SF of T will be equal to 25 multiplied by parenthesis of 1 minus e to the power of negative T divided by 5, and don't forget that our units are in meters, which that's what M denotes. And that's it. That is our second answer. Hooray, we did it. So now, our 4th step is we need to focus on longtime behavior now. So with that in mind, as we should recall and note that for Emily's case, let's start with Emily, which I'll just say E for short. So for Emily's case, The limit as T approaches positive infinity, so the limit as T approaches positive infinity of SE of T will be equal to the limit, so be equal to the limit. As T approaches positive infinity of 5 multiplied by the natural log of T + 3 divided by 3 will be equal to infinity. So that will mean, as a result, Emily travels an unbounded infinite distance as T approaches zero, whereas for Frank, Frank will represent well, Frank's case, we can say that the limit as T approaches positive infinity. Of S of F of T. So SF of T will be equal to the limit as T approaches positive infinity of 25 multiplied by parentheses 1 minus E to the power of negative T divided by 5, which will be equal to simply 25. So, as a result, Frank travels only a finite total distance of 25 m. And so that will mean that our final answer, our third and final answer will be Frank, because Frank is the only one who travels a finite distance based on the information we've collected together. Hooray, we did it. So going back to the top to recap what we've learned, that will mean that our final three answers once again will be as follows. So, SE of T, so the distance, the position function for Emily, SEF T will be equal to 5 multiplied by the natural log of T + 3 divided by 3 m. And for Frank's position function, SF of T will be equal to 25 multiplied by parenthesis of 1 minus e to the power of negative T divided by 5 m. And finally, our third and final answer will be Frank, because Frank is the only one out of the two walkers that travels a finite distance. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Key Concepts

Velocity and Position Relationship

Definite and Indefinite Integration

Limits and Infinite Time Behavior

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