Question

At noon ​(t=​0), Alicia starts walking along a​ long, straight road at 4 ​mi/hr. Her velocity decreases according to the function ​v(t)=4​/(t+1​) for t≥0. At​ noon, Boris also starts walking along the same road with a 3​-mi head start on​ Alicia; his velocity is given by ​u(t)=2​/(t+1​) for t≥0. a. Find the position functions for Alicia and​ Boris, where s=0 corresponds to​ Alicia's starting point. b.​ When, if​ ever, does Alicia overtake​ Boris?

Answer 1

The position function for Alicia is s(t) = 4ln(t+1), and for Boris it is r(t) = 2ln(t+1) + 3. Alicia overtakes Boris at around t ≈ 3.482 hours.

Step-by-step explanation:

To find the position functions for Alicia and Boris, we need to integrate their respective velocity functions over time. For Alicia, the velocity function is v(t) = 4/(t+1).

Integrating this gives us the position function s(t) = 4ln(t+1) + C, where C is a constant of integration. Since Alicia starts at s = 0 at t = 0, we have C = 0, so the position function for Alicia is s(t) = 4ln(t+1).

For Boris, his velocity function is u(t) = 2/(t+1). Integrating this gives us the position function r(t) = 2ln(t+1) + D, where D is a constant of integration. Since Boris starts with a 3-mi head start, we have r(0) = 3.

Plugging this into the position function gives us D = 3, so the position function for Boris is r(t) = 2ln(t+1) + 3.

To determine when Alicia overtakes Boris, we need to find the time t such that s(t) = r(t). Setting 4ln(t+1) = 2ln(t+1) + 3 and solving for t gives us t = e^(3/2) - 1 ≈ 3.482.

Therefore, Alicia overtakes Boris at around t ≈ 3.482 hours.