A ballon is rising vertically above a level, straight road at a constant rate of

Question

A ballon is rising vertically above a level, straight road at a constant rate of 1ft/sec. Just when the ballon is 65 ft above the ground, a bicycle moving at a constant rate of 17ft/sec passes under it. How fast is the distance s(t) between the bicycle and ballon increasting 3 sec later? I know I have to derivate using Pythagorean theorem: dh/dt =1 h=64 ft dx/dt = 17 but I don't understand how to get these values: at t=3 y(t)=68 at t=3 x(t)=51 at t=3 s(t) = 85 *I can see for myself the answer in the textbook solution, so please explain the process...WILL RATE! :)

Explanation / Answer

At 3 seconds the balloon is 68 ft above the air bc 65 +3(1)=68. T(3)y(t)=68 At 3 seconds the bike has traveled 51ft bc 17*3=51 t(3) y(t)=51 A^2 +b^2 =c^2 68*68+51*51=c^2=7225 Sqrt(7225)=85=c

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