Question

At noon, you are running to get to class and notice a friend 100 feet west of you, also running to class. If you are running south at a constant rate of 450 ft/min (approximately 5 mph) and your friend is running north at a constant rate of 350 ft/min (approximately 4 mph), how fast is the distance between you and your friend changing at 12:02 pm?

Answer 1

Explanation:

Velocity of first Person
`v_1=450 ft/min`

Velocity of second Person
`v_2=350 ft/min`

Distance between them P=100 ft

Let x be the distance moved by 1 st and y be the distance moved by second person by the time 12:02 PM

In Triangle ABC

`BC=350 t+450 t`

`BC=800 t`

using Pythagoras

`AC^2=AB^2+BC^2`

`AC^2=100^2+(800t)^2`

differentiating with respect to time

`2* AC* \frac{\mathrm{d} AC}{\mathrm{d} t}=2* (800t)* 800`

`\frac{\mathrm{d} AC}{\mathrm{d} t}=(800^2* t)/(AC)`

AC after 2 minute

`AC=1603.121 ft`

`\frac{\mathrm{d} AC}{\mathrm{d} t}=(800^2* 2)/(1603.121)`

`\frac{\mathrm{d} AC}{\mathrm{d} t}=(64* 10^4* 2)/(1603.121)`

`\frac{\mathrm{d} AC}{\mathrm{d} t}=798.44 ft/min`