Answer:
16,126 ft
Step-by-step explanation:
Using the net change theorem and letting s(t) represent the aircraft's position, and s(0) be the aircraft's position at time t = 0, s(0) = 15000 ft and s(15 be the aircraft's position at time, t = 15 minutes respectively, then,
s(15) - s(0) = ∫₀¹⁵f(t)dt
s(15) - s(0) = ∫₀¹⁵[[tex]e^{0.4t} + 8[/tex]]dt
s(15) - 15000 = [[tex]\frac{e^{0.4t}}{0.4} + 8t[/tex]]₀¹⁵
s(15) - 15,000 ft = [tex]\frac{e^{0.4X15}}{0.4} + 8X15 - \frac{e^{0.4 X 0}}{0.4} + 8(0)\\\frac{e^{6}}{0.4} + 120 - \frac{e^{0}}{0.4} + 0\\\frac{403.43}{0.4} + 120 - \frac{1}{0.4} + 0\\1008.57 + 120 - 2.5\\1126.07[/tex]
s(15) = 15,000 ft + 1126.07 ft
s(15) = 16,126.07 ft
s(15) ≅ 16,126 ft to the nearest foot
