draw the figure, a right triangle. figure out distances after 20seconds.
let horizontal distance be x, height be h, and d the slant distance.
d^2=h^2+x^2
2d dd/dt= 2h dh/dt+2x dx/dt
solve for dd/dt, you know all else.
A hot-air balloon is 180 ft above the ground when a motorcycle passes directly beneath it (travelling in a straight line on a horizontal road) going at a constant speed of 60 ft/s. If the balloon is rising vertically at a rate of 15 ft/s, what is the rate of change of the distance between the motorcycle and the balloon 20 seconds after the motorcycle was directly beneath the balloon.
2 answers
Wow, this is such an old question. The answer is easy, namely:
during the 20 seconds that passed:
- baloon increased altitude by 300ft,
- car drove 1200ft and so:
Initial distance between them was 180ft and after 20 seconds
it s squareRoot(power(480,2) + power(1200,2)), which is 1292,44ft.
Distance changed by 1112,44ft
during the 20 seconds that passed:
- baloon increased altitude by 300ft,
- car drove 1200ft and so:
Initial distance between them was 180ft and after 20 seconds
it s squareRoot(power(480,2) + power(1200,2)), which is 1292,44ft.
Distance changed by 1112,44ft
0 answers