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a man starts walking north at 45 m/min from a point p. five minutes later a woman starts walking south at 100 m/min from a point 700 m due east of p. at what rate are they separating 15 minutes after the woman starts ?

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2 Answers

The answer will be somewhat analogous to the question you previously asked. This time, the distance between the couple, $z$, can be expressed as $z=\sqrt{700^2+x^2}$, where $x=x_1+x_2$, and $x_1$ is the distance of man from his starting point and $x_2$ is the distance of the woman from her starting point (we can simply add the distances because they are moving in opposite directions). Now express the distance walked as a function of time, and you get $x_1=225+45t$ and $x_2=100t$. Finally, take the derivative of $z(t)$ wrt to $t$, and set $t=15$.

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Thanks for the help :) –  mimzy Mar 29 '13 at 21:15
    
You're welcome :) –  Johnny Westerling Mar 29 '13 at 21:16
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Let us start from graphing the problem.

graph

Here Q is the point from where woman starts walking, A and B are the points where man and woman are at the moment of time $t$. Hence $x=x(t)$ is the distance traveled by man from the very beginning, and $y=y(t)$ is the distance traveled by woman at the same timespan.
Now let us find a formula for the distance between man and woman $$S(t)=|AO|+|OB|=\sqrt{x^2+z^2}+\sqrt{y^2+(R-z)^2}$$ But what about $z$? Here we can use the fact that triangles APO and OQB are similar, hence $$\frac{z}{R-z}=\frac{x}{y}, z=\frac{Rx}{x+y}$$ Now substitute $z$ into $S(t)$, differentiate it and get the value in desired point. I left the calculations to you.

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Thanks for the help :) –  mimzy Mar 29 '13 at 21:14
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