Question

A​ hot-air balloon is 140 ft140 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 hr60 mi/hr ​(88 ft divided by s88 ft/s​). If the balloon rises vertically at a rate of 14 ft divided by s14 ft/s​, what is the rate of change of the distance between the motorcycle and the balloon 10 seconds10 seconds ​later?

Answer 1

Answer:

Step-by-step explanation:

Given

Balloon velocity is 14 ft/s upwards  

Distance between balloon and cyclist is 140 ft

Velocity of motor cycle is 88 ft/s

After 10 sec  

motorcyclist traveled a distance of $d_c=88\times 10=880 ft$

Distance traveled by balloon in 10 s

$d_b=14\times 10=140 ft$

net height of balloon from ground =140+140=280 ft[/tex]

at $t=10 s$

distance between cyclist and balloon is $z=\sqrt{280^2+880^2}$

$z=923.47 ft$

now suppose at any time t motorcyclist cover a distance of x m and balloon is at a height of h m

thus $z^2=(88t)^2+(14t+140)^2$

differentiating w.r.t time

$\Rightarrow z\frac{\mathrm{d} z}{\mathrm{d} t}=14\cdot \left ( 14t+140\right )+88\cdot \left ( 88t\right )$

$\Rightarrow at\ t=10 s$

$\Rightarrow \frac{\mathrm{d} z}{\mathrm{d} t}=\frac{14\cdot \left ( 14t+140\right )+88\cdot \left ( 88t\right )}{z}$

$\Rightarrow \frac{\mathrm{d} z}{\mathrm{d} t}=\frac{14\times 280+88^2\times 10}{923.471}$

$\Rightarrow \frac{\mathrm{d} z}{\mathrm{d} t}=\frac{81,360}{923.471}=88.102\ ft/s$