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 find the distance the train travels between \( t=0.1 \) and \( t=0.3 \), we first need to find the velocity function. We can integrate the acceleration function to get the velocity. Given that the acceleration \( a(t) = -1120(1 + 4t)^{-3} \), we integrate with respect to \( t \): \[ v(t) = \int a(t) \, dt = \int -1120(1 + 4t)^{-3} \, dt \] Using substitution \( u = 1 + 4t \) gives \( du = 4 \, dt \) or \( dt = \frac{du}{4} \). The limits of integration will change accordingly: If \( t = 0.1 \), \( u = 1 + 4(0.1) = 1.4 \). If \( t = 0.3 \), \( u = 1 + 4(0.3) = 1.2 \). Thus, the integral becomes: \[ v(t) = -\frac{1120}{4} \int (u)^{-3} \, du = -280\left[-\frac{1}{2u^{2}}\right] + C = \frac{140}{u^{2}} + C \] Now, we apply the initial condition. At \( t = 0 \), we have \( v(0) = 140 \): \[ v(0) = \frac{140}{(1)^2} + C = 140 \implies C = 0 \] So, the velocity function is: \[ v(t) = \frac{140}{(1 + 4t)^2} \] Next, we find the distance traveled between \( t=0.1 \) and \( t=0.3 \): \[ d = \int_{0.1}^{0.3} v(t) \, dt = \int_{0.1}^{0.3} \frac{140}{(1 + 4t)^2} \, dt \] Using the substitution \( u = 1 + 4t \): When \( t = 0.1 \), \( u = 1.4 \) and when \( t = 0.3 \), \( u = 1.2 \). The differential becomes \( dt = \frac{du}{4} \): \[ d = \int_{1.4}^{1.2} \frac{140}{u^2} \cdot \frac{1}{4} \, du = 35 \int_{1.4}^{1.2} u^{-2} \, du = 35 \left[-\frac{1}{u}\right]_{1.4}^{1.2} = 35 \left(-\frac{1}{1.2} + \frac{1}{1.4}\right) \] Calculating the above term: \[ = 35 \left(-\frac{10}{12} + \frac{12}{14}\right) = 35 \left(-\frac{5}{6} + \frac{6}{7}\right) = 35 \left(-\frac{35}{42} + \frac{36}{42}\right) = 35 \left(\frac{1}{42}\right) = \frac{35}{42} \approx 0.833 \text{ mi} \] Thus, between \( t=0.1 \) and \( t=0.3 \), the train travels approximately \( 0.833 \) mi.
