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

A slowing race Starting at the same time and place, Abe and Bob race, running at velocities $u(t)=4 /(t+1) \mathrm{mi} / \mathrm{hr}$ and $v(t)=4 e^{-t / 2} \mathrm{mi} / \mathrm{hr},$ respectively, for $t \geq 0$ a. Who is ahead after $t=5$ hr? After $t=10 \mathrm{hr}$ ? b. Find and graph the position functions of both runners. Which runner can run only a finite distance in an unlimited amount of time?

A slowing race Starting at the same time and place, Abe and Bob race, running at velocities $u(t)=4 /(t+1) \mathrm{mi} / \mathrm{hr}$ and $v(t)=4 e^{-t / 2} \mathrm{mi} / \mathrm{hr},$ respectively, for $t \geq 0$
a. Who is ahead after $t=5$ hr? After $t=10 \mathrm{hr}$ ?
b. Find and graph the position functions of both runners. Which runner can run only a finite distance in an unlimited amount of time?

Key Concepts

Integration of Specific Functions

Understanding the integration of functions like a reciprocal function (leading to logarithmic functions) and an exponential function (leading to expressions involving exponential decay) is important. These integrations help obtain closed-form position functions that describe the runners’ movements over time.

Convergence of Improper Integrals

The concept of convergence in the context of improper integrals is used to determine whether an object traveling in motion under diminishing speed covers a finite or infinite distance over an unlimited amount of time. This idea is applied to conclude which runner only covers a finite distance as time approaches infinity.

Position Function from Velocity

In calculus, the position function of an object is found by integrating its velocity function over time. This concept is crucial because it quantifies the total displacement from the starting point, including how the object’s speed at each instant contributes to the overall distance covered.

Definite Integration

Definite integration is used to compute the exact displacement between two specific times by integrating the velocity function between those time limits. This method is essential for comparing positions, such as determining which runner is ahead at a given moment in time.

Transcript

00:01 We're given functions for the velocity of two runners.

00:04 The first is abe represented by a of t, and the second is bob represented by b of t.

00:11 And we want to know who's ahead when t is 5 and who's ahead when t is 10.

00:15 And then we want to graph these two position functions that we find and see which one is bounded and thus which velocity will be finite.

00:24 So we'll start with part a.

00:25 The position, if we have the velocity, the position is given by the integral of velocity.

00:40 So when we want abe's position at 5 of t for t, we'll take the integral from 0 to 5, so the initial to what we're asked about.

00:56 And then we'll put in our a function, 4 over t plus 1 d t.

01:02 The 4 can come out.

01:03 We can recognize that this is 1 over a function, and so we'll get the natural log.

01:09 So it'll be 4lnt plus 1.

01:13 When we take the integral, this is evaluated from 0 to 5.

01:18 The natural log of 1 is 0, so we're only going to maintain our 4ln of 5 plus 1 is 6.

01:27 We put that into our calculator, we'll get 7 .17 miles is the position.

01:33 Now we'll do the same for b.

01:35 The integral from 0 to 5 of our bob function is 4e.

01:41 It's the negative t over 2 d t.

01:46 Now this 4 can come out as a constant and the negative t over 2 will get negative 2 multiplied by our outcome.

01:56 And so we'll look like this.

01:59 To the negative t over 2.

02:01 That's evaluated from 0 to 5.

02:04 We don't want that negative in there, so we'll swap the bounds.

02:08 That'll be positive 8e to the negative t over 2 from 5 to 0 to get rid of that negative value there.

02:17 We do this out, we'll get e to 0 is 1, and so this will be 8 minus.

02:24 When we put in 5 for t, we get 8 e to the negative 5 over 2.

02:29 We can use a calculator for this, and get 7 .34 miles as the position, and that's greater than 7 .17, so bob is ahead after t is 5.

02:43 Now we need to do the same for 10, so a10 is equal to the integral from 0 to 10 of our abe function, 4 over t plus 1, dt...

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