At noon \( (t=0) \), Alicia starts walking along a long, straight road at \( 5 \mathrm{mi} / \mathrm{hr} \). Her velocity decreases according to the function \( v(t)=5 /(t+1) \) for \( t \geq 0 \). At noon, Boris also starts walking along the same road with a 1 -mi head start on Alicia; his velocity is given by \( u(t)=3 /(t+1) \) for \( t \geq 0 \). 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?
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To find the position functions for Alicia and Boris, let's first derive their position from their velocities. ### Part a: Alicia's velocity function is \( v(t) = \frac{5}{t+1} \). To find her position function \( s_A(t) \), we need to integrate the velocity function: \[ s_A(t) = \int v(t) \, dt = \int \frac{5}{t+1} \, dt = 5 \ln(t+1) + C_A. \] Since Alicia starts at \( s_A(0) = 0 \), we can set \( C_A = 0 \). Thus, Alicia's position function is: \[ s_A(t) = 5 \ln(t+1). \] Boris has a 1-mile head start, so his position function (starting at 1 mile) is based on his velocity \( u(t) = \frac{3}{t+1} \): \[ s_B(t) = \int u(t) \, dt = \int \frac{3}{t+1} \, dt = 3 \ln(t+1) + C_B. \] With the initial position at \( s_B(0) = 1 \), we find \( C_B = 1 \). Thus, Boris' position function is: \[ s_B(t) = 3 \ln(t+1) + 1. \] ### Part b: To determine when Alicia overtakes Boris, we need to find \( t \) such that: \[ s_A(t) = s_B(t). \] This leads to: \[ 5 \ln(t+1) = 3 \ln(t+1) + 1. \] Rearranging gives: \[ 5 \ln(t+1) - 3 \ln(t+1) = 1, \] \[ 2 \ln(t+1) = 1. \] Dividing both sides by 2 yields: \[ \ln(t+1) = \frac{1}{2}. \] Exponentiating both sides results in: \[ t+1 = e^{1/2}, \] so \[ t = e^{1/2} - 1 \approx 0.6487. \] Thus, Alicia will overtake Boris approximately 0.6487 hours after noon, or around 38.92 minutes after noon.
