The velocity of an airplane flying into a headwind is given by \( v(t)=30\left(16-t^{2}\right) \mathrm{mi} / \mathrm{hr} \) for \( 0 \leq t \leq 3 \mathrm{hr} \). Assume that \( \mathrm{s}(0)=0 \). c. When the velocity reaches 350 mi/hr, the airplane has traveled \( \square \mathrm{mi} \). (Do not round until the final answer. Then round to two decimal places as needed.)

We are given the airplane’s velocity function   v(t) = 30(16 – t²)  (in miles per hour), for 0 ≤ t ≤ 3 hours and are told that s(0) = 0. We want to find the distance traveled when the airplane’s velocity reaches 350 mi/hr. Step 1. Find the time t when v(t) = 350. Set up the equation:   30(16 – t²) = 350 Divide both sides by 30:   16 – t² = 350⁄30 = 35⁄3 Solve for t²:   t² = 16 – (35⁄3)     = (48⁄3 – 35⁄3) = 13⁄3 Thus,   t = √(13⁄3) Step 2. Find the displacement s(t) when t = √(13⁄3). Since s(0) = 0, the displacement is the integral of the velocity:   s(t) = ∫₀ᵗ v(u) du = ∫₀ᵗ 30(16 – u²) du Compute the integral:   s(t) = 30 [16u – u³⁄3] evaluated from 0 to t     = 30 (16t – t³⁄3) Substitute t = √(13⁄3):   s(√(13⁄3)) = 30 [16√(13⁄3) – ( (√(13⁄3))³⁄3 )] Notice that   (√(13⁄3))³ = (13⁄3) √(13⁄3) Thus,   s(√(13⁄3)) = 30 [16√(13⁄3) – (1/3)((13⁄3)√(13⁄3))]     = 30 [√(13⁄3) (16 – 13⁄9)] Write 16 as 144⁄9:   16 – 13⁄9 = (144 – 13)⁄9 = 131⁄9 So,   s(√(13⁄3)) = 30 · √(13⁄3) · (131⁄9)     = (30·131)/(9) · √(13⁄3) We can simplify 30/9 to 10/3:   s(√(13⁄3)) = (10·131)/3 · √(13⁄3)     = (1310/3) · √(13/3) Step 3. Compute a numerical value. First, compute √(13/3):   13/3 ≈ 4.33333, so √(4.33333) ≈ 2.0801 Next, compute (1310/3):   1310/3 ≈ 436.6667 Now, multiply:   s ≈ 436.6667 × 2.0801 ≈ 908.30 miles Final Answer: When the velocity reaches 350 mi/hr, the airplane has traveled approximately 908.30 miles.