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.)

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**.