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

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* 10=880 ft`

Distance traveled by balloon in 10 s

`d_b=14* 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=√(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}=(14\cdot \left ( 14t+140\right )+88\cdot \left ( 88t\right ))/(z)`

`\Rightarrow \frac{\mathrm{d} z}{\mathrm{d} t}=(14* 280+88^2* 10)/(923.471)`

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