The velocity of an airplane flying into a headwind is given by \( \mathrm{v}(\mathrm{t})=30\left(16-\mathrm{t}^{2}\right) \mathrm{mi} / \mathrm{hr} \) for \( 0 \leq \mathrm{t} \leq 3 \mathrm{hr} \). Assume that \( \mathrm{s}(0)=0 \). a. Determine the position function for \( 0 \leq t \leq 3 \). b. How far does the airplane travel in the first 2 hr ? c. How far has the airplane traveled at the instant its velocity reaches \( 450 \mathrm{mi} / \mathrm{hr} \) ?

To find the position function and answer the subsequent questions about the airplane's travel, we'll start with the velocity function given. ### Part a: Determine the position function The position function \( s(t) \) can be found by integrating the velocity function \( v(t) \): \[ s(t) = \int v(t) \, dt = \int 30(16 - t^2) \, dt \] Calculating the integral: \[ s(t) = 30 \left( 16t - \frac{t^3}{3} \right) + C \] Since the airplane starts from rest at \( s(0) = 0 \): \[ s(0) = 30 \left( 16 \cdot 0 - \frac{0^3}{3} \right) + C = 0 \implies C = 0 \] Thus, the position function is: \[ s(t) = 480t - 10t^3 \quad \text{for } 0 \leq t \leq 3 \] ### Part b: Distance traveled in the first 2 hours Now we’ll evaluate the position function at \( t = 2 \) hours: \[ s(2) = 480(2) - 10(2^3) = 960 - 80 = 880 \text{ miles} \] Thus, the airplane travels **880 miles** in the first 2 hours. ### Part c: Distance at the instant velocity reaches 450 mi/hr Set the velocity function equal to 450 and solve for \( t \): \[ 30(16 - t^2) = 450 \] Dividing both sides by 30: \[ 16 - t^2 = 15 \implies t^2 = 1 \implies t = 1 \text{ hour} \] Now, substitute \( t = 1 \) into the position function to find the distance traveled: \[ s(1) = 480(1) - 10(1^3) = 480 - 10 = 470 \text{ miles} \] Thus, the airplane has traveled **470 miles** at the instant its velocity reaches \( 450 \text{ mi/hr} \).