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

Answer:

Step-by-step 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

$2\times AC\times \frac{\mathrm{d} AC}{\mathrm{d} t}=2\times (800t)\times 800$

$\frac{\mathrm{d} AC}{\mathrm{d} t}=\frac{800^2\times t}{AC}$

AC after 2 minute

$AC=1603.121 ft$

$\frac{\mathrm{d} AC}{\mathrm{d} t}=\frac{800^2\times 2}{1603.121}$

$\frac{\mathrm{d} AC}{\mathrm{d} t}=\frac{64\times 10^4\times 2}{1603.121}$

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