Let t sec be the time since the balloon was directly above the bike, and let d ft be the distance between the balloon and the bike.
According to my diagram,
d^2 = (58t)^2 + (150 + 10t)^2
= 3364t^2 + 22500 + 3000t + 100t^2
= 3464t^2 + 3000t
2d dd/dt = 6928t + 3000
dd/dt = (3464t + 1500)/d
plug t = 10 into the d^2 = ... equation, and find d
now plut t = 10, and d = ....
into the dd/dt expression , remember the units would be ft/s
A hot air balloon is 150 ft above the ground when a motorcycle passes beneath it (traveling in a striaght line on a horizontal road) going 58 ft/sec. If the balloon is rising vertically at a rate of 10 ft/sec, what is the rate of change of the distance between the motorcycle and the balloon 10 seconds later?
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