Answer:
Step-by-step explanation:
Velocity of first Person [tex] v_1=450 ft/min[/tex]
Velocity of second Person [tex]v_2=350 ft/min[/tex]
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
[tex]BC=350 t+450 t[/tex]
[tex]BC=800 t [/tex]
using Pythagoras
[tex]AC^2=AB^2+BC^2[/tex]
[tex]AC^2=100^2+(800t)^2[/tex]
differentiating with respect to time
[tex]2\times AC\times \frac{\mathrm{d} AC}{\mathrm{d} t}=2\times (800t)\times 800[/tex]
[tex]\frac{\mathrm{d} AC}{\mathrm{d} t}=\frac{800^2\times t}{AC}[/tex]
AC after 2 minute
[tex]AC=1603.121 ft[/tex]
[tex]\frac{\mathrm{d} AC}{\mathrm{d} t}=\frac{800^2\times 2}{1603.121}[/tex]
[tex]\frac{\mathrm{d} AC}{\mathrm{d} t}=\frac{64\times 10^4\times 2}{1603.121}[/tex]
[tex]\frac{\mathrm{d} AC}{\mathrm{d} t}=798.44 ft/min[/tex]
