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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**Problem 1(a): Find the position functions for Alicia and Boris, where $$ s=0 $$ corresponds to Alicia's starting point.**

**Solution:**

To determine the position functions for Alicia and Boris, we'll integrate their velocity functions over time, considering their initial positions.

1. **Alicia's Position Function ($$ s_A(t) $$):**
   
   - **Velocity Function:** $$ v(t) = \frac{5}{t + 1} $$ mi/hr
   - **Initial Position:** $$ s_A(0) = 0 $$ mi (since $$ s=0 $$ is Alicia's starting point)

**Integration:**
   $$
   s_A(t) = \int_{0}^{t} v(\tau) \, d\tau = \int_{0}^{t} \frac{5}{\tau + 1} \, d\tau = 5 \ln(\tau + 1) \Big|_{0}^{t} = 5 \ln(t + 1)
   $$

2. **Boris's Position Function ($$ s_B(t) $$):**
   
   - **Velocity Function:** $$ u(t) = \frac{3}{t + 1} $$ mi/hr
   - **Initial Position:** $$ s_B(0) = 1 $$ mi (Boris has a 1-mile head start)

**Integration:**
   $$
   s_B(t) = s_B(0) + \int_{0}^{t} u(\tau) \, d\tau = 1 + \int_{0}^{t} \frac{3}{\tau + 1} \, d\tau = 1 + 3 \ln(\tau + 1) \Big|_{0}^{t} = 1 + 3 \ln(t + 1)
   $$

**Final Position Functions:**
$$
\boxed{
\begin{aligned}
s_A(t) &= 5 \ln(t + 1) \
s_B(t) &= 1 + 3 \ln(t + 1)
\end{aligned}
}
$$

Problem 1 (a) Answer:

After simplifying, the position functions are

s_A(t) = 5 ln(t + 1)

and

s_B(t) = 1 + 3 ln(t + 1).

In symbols,

s_A(t) = 5 ln(t+1)

s_B(t) = 1 + 3 ln(t+1).
**Problem 1(a): Position Functions** - **Alicia's Position:** $$ s_A(t) = 5 \ln(t + 1) $$ miles - **Boris's Position:** $$ s_B(t) = 1 + 3 \ln(t + 1) $$ miles **Problem 1(b): When Does Alicia Overtake Boris?** To find when Alicia overtakes Boris, set $$ s_A(t) = s_B(t) $$: $$ 5 \ln(t + 1) = 1 + 3 \ln(t + 1) $$ Subtract $$ 3 \ln(t + 1) $$ from both sides: $$ 2 \ln(t + 1) = 1 $$ Divide both sides by 2: $$ \ln(t + 1) = \frac{1}{2} $$ Exponentiate both sides to solve for $$ t $$: $$ t + 1 = e^{\frac{1}{2}} \approx 1.6487 $$ Subtract 1 from both sides: $$ t \approx 0.6487 \text{ hours} $$ Convert hours to minutes: $$ 0.6487 \times 60 \approx 38.92 \text{ minutes} $$ **Conclusion:** Alicia overtakes Boris approximately 38.92 minutes after noon. **Simplified Answer:** Alicia overtakes Boris about 39 minutes after noon.

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