Question

a hot - air balloon is 180 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going 60 mi/hr (88 ft/s). if the balloon rises vertically at a rate of 14 ft/s, what is the rate of change of the distance between the motorcycle and the balloon 9 seconds later? the rate of change of the distance between the motorcycle and the balloon after 9 seconds is about (round to two decimal places as needed.)

Explanation:

Step1: Find the horizontal distance of the motorcycle

The speed of the motorcycle is $v = 88$ ft/s. Using the formula $d=v\times t$, where $t = 9$ s. So the horizontal distance $x=88\times9 = 792$ ft.

Step2: Find the vertical distance of the balloon

The initial height of the balloon is $h_0=180$ ft and it rises at a rate of $r = 14$ ft/s. Using the formula $h=h_0+r\times t$, with $t = 9$ s, we get $h=180 + 14\times9=180+126 = 306$ ft.

Step3: Define the distance formula

Let $D$ be the distance between the motorcycle and the balloon. By the Pythagorean theorem, $D=\sqrt{x^{2}+h^{2}}$.

Step4: Differentiate the distance formula with respect to time $t$

We have $D^{2}=x^{2}+h^{2}$. Differentiating both sides with respect to $t$ gives $2D\frac{dD}{dt}=2x\frac{dx}{dt}+2h\frac{dh}{dt}$, so $\frac{dD}{dt}=\frac{x\frac{dx}{dt}+h\frac{dh}{dt}}{D}$.

Step5: Calculate the values of $D$, $\frac{dx}{dt}$, and $\frac{dh}{dt}$

We know that $\frac{dx}{dt}=88$ ft/s, $\frac{dh}{dt}=14$ ft/s, $x = 792$ ft, $h = 306$ ft. Then $D=\sqrt{792^{2}+306^{2}}=\sqrt{627264 + 93636}=\sqrt{720900}=849$ ft.

Step6: Calculate $\frac{dD}{dt}$

Substitute the values into the formula for $\frac{dD}{dt}$: $\frac{dD}{dt}=\frac{792\times88+306\times14}{849}=\frac{69696+4284}{849}=\frac{73980}{849}\approx87.14$ ft/s.

Answer:

$87.14$