Question

A​ hot-air balloon is 150 ft above the ground when a motorcycle​ (traveling in a straight line on a horizontal​ road) passes directly beneath it going 60 mi divided by hr ​(88 ft divided by s​). If the balloon rises vertically at a rate of 8 ft divided by s​, what is the rate of change of the distance between the motorcycle and the balloon 10 seconds ​later?

Answer 1

Answer:

 dR/dt = 10.2 ft / s

Explanation:

Let's work this problem by finding the distance between the balloon and the motorcycle and then drift for the speed change of the distance

Balloon

      y = y₀ +$v_{oy}$ t

Motorcycle

      x = v₀ₓ t

Distance, let's use Pythagoras' theorem

      R² = x² + y²

      R² = (v₀ₓ t)² + (y₀ + $v_{oy}$ t)²

     v₀ₓ = 88 ft / s

     $v_{oy}$ = 8 ft / s

     y₀ = 150 ft

     R² = (8 t)² + (150 + 8 t )²

     R² = 64 t² + (150 + 8t )²

This is the expression for the distance between the two bodies, the rate of change is the derivative with respect to time (d / dt)

         2RdR / dt = 64 2 t + 2 (150 + 8t) 8

        dR / dt = [64 t + (1200 + 64t )] / R

dR/dt = (1200 +128 t)/R

Let's calculate for the time of 10 s

        dR / dt = (1200 + 128 10) / R = 2480 /R

       R = √ [64 10² + (150 + 8 10)²

       R = √ [6400 + 52900]

       R = 243.5 ft

       dR / dt = (2480) / 243.5

       dR / dt = 10.2 ft / s