At $$ t=0 $$, a train approaching a station begins decelerating from a speed of $$ 140 \mathrm{mi} / \mathrm{hr} $$ according to the acceleration function $$ \mathrm{a}(\mathrm{t})=-1120(1+4 \mathrm{t})^{-3}{\mathrm{mi} / \mathrm{hr}^{2}}^{2} $$, where $$ t \geq 0 $$. How far does the train travel between $$ \mathrm{t}=0 $$ and $$ \mathrm{t}=0.1 $$ ? Between $$ \mathrm{t}=0.1 $$ and $$ \mathrm{t}=0.3 $$ ? Between $$ \mathrm{t}=0 $$ and $$ \mathrm{t}=0.1 $$ the train travels 10.000 mi . (Round to three decimal places as needed.) Between $$ \mathrm{t}=0.1 $$ and $$ \mathrm{t}=0.3 $$ the train travels $$ \square \mathrm{mi} $$. (Round to three decimal places as needed.)

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To determine the distance the train travels between $$ t = 0.1 $$ hours and $$ t = 0.3 $$ hours, we'll follow these steps:

1. **Determine the Velocity Function $$ v(t) $$:**

The acceleration function is given by:
   $$
   a(t) = \frac{dv}{dt} = -1120(1 + 4t)^{-3} \ \text{mi/hr}^2
   $$
   
   Integrating the acceleration to find the velocity:
   $$
   v(t) = \int a(t) \, dt + v_0
   $$
   
   Let $$ u = 1 + 4t $$, then $$ du = 4 \, dt $$ or $$ dt = \frac{du}{4} $$:
   $$
   v(t) = -1120 \int u^{-3} \cdot \frac{du}{4} + v_0 = -280 \int u^{-3} \, du + v_0 = 140 u^{-2} + C
   $$
   
   Applying the initial condition $$ v(0) = 140 \ \text{mi/hr} $$:
   $$
   140 = 140(1)^{-2} + C \implies C = 0
   $$
   
   Thus, the velocity function is:
   $$
   v(t) = \frac{140}{(1 + 4t)^2} \ \text{mi/hr}
   $$

2. **Determine the Position Function $$ x(t) $$:**

Integrate the velocity to find the position:
   $$
   x(t) = \int v(t) \, dt + x_0
   $$
   
   Using the previously defined substitution $$ u = 1 + 4t $$:
   $$
   x(t) = \int \frac{140}{u^2} \cdot \frac{du}{4} = 35 \int u^{-2} \, du = -35u^{-1} + C
   $$
   
   Applying the initial condition $$ x(0) = 0 $$:
   $$
   0 = -35(1)^{-1} + C \implies C = 35
   $$
   
   Thus, the position function is:
   $$
   x(t) = 35\left(1 - \frac{1}{1 + 4t}\right)
   $$

3. **Calculate the Distance Between $$ t = 0.1 $$ and $$ t = 0.3 $$:**

Evaluate $$ x(t) $$ at $$ t = 0.3 $$:
   $$
   x(0.3) = 35\left(1 - \frac{1}{1 + 4(0.3)}\right) = 35\left(1 - \frac{1}{2.2}\right) \approx 35 \times 0.54545 \approx 19.0909 \ \text{mi}
   $$
   
   The distance traveled between $$ t = 0.1 $$ and $$ t = 0.3 $$ is:
   $$
   x(0.3) - x(0.1) = 19.0909 \ \text{mi} - 10.000 \ \text{mi} = 9.0909 \ \text{mi}
   $$
   
   Rounded to three decimal places:
   $$
   9.091 \ \text{mi}
   $$

**Answer:**  
Between $$ \mathrm{t}=0.1 $$ and $$ \mathrm{t}=0.3 $$ the train travels **9.091 mi**.
Between $$ t=0.1 $$ and $$ t=0.3 $$, the train travels **9.091 miles**.

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