relative rate of moving problem

[asifrahman1988]

A man starts walking north at 1.2 m/s from a point P. Five minutes later a woman starts walking south at 1.6 m/s from a point 200 m due east of P. At what rate are the peoplemoving apart 15 min after the woman starts walking?

 

[Deleted member 4993]

A man starts walking north at 1.2 m/s from a point P. Five minutes later a woman starts walking south at 1.6 m/s from a point 200 m due east of P. At what rate are the peoplemoving apart 15 min after the woman starts walking?

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[soroban]

Hello, asifrahman1988!

A man starts walking north at 1.2 m/s from a point P.
Five minutes later a woman starts walking south at 1.6 m/s from a point Q 200 m due east of P.
At what rate are the people moving apart 15 min after the woman starts walking?

    M *
      |\
 1.2t | \
      |  \
    A *   \
      |    \
  360 |     \ x
      |      \
    P *-------\---* Q
      :        \  |
 1.6t :         \ | 1.6t
      :          \|
    B * - - - - - * W
           200

The man starts at \(\displaystyle P\).
In 5 minutes (300 seconds) he walks to \(\displaystyle A.\)
. . \(\displaystyle PA \,=\,(300)(1.2) \,=\,360\text{ m}\)
In the next \(\displaystyle t\) seconds, he walks \(\displaystyle 1.2t\) m to point \(\displaystyle M.\)

In the same \(\displaystyle t\) seconds, the woman walks \(\displaystyle 1.6t\) m from point \(\displaystyle Q\) to point \(\displaystyle W.\)

Extend \(\displaystyle MP\) downward to meet the horizontal line through \(\displaystyle W\) at \(\displaystyle B.\)
. . \(\displaystyle PQ \,=\,BW\,=\,200\text{ m}\)
Let \(\displaystyle x \,=\,MW.\)


Pythagorus says: .\(\displaystyle x^2 \;=\;M\!B^2 + BW^2\)

We have: .\(\displaystyle x^2 \:=\: (2.8t + 360)^2 + 200^2 \:=\:7.84t^2 + 2016t + 169,\!600 \)

Hence: . \(\displaystyle x \:=\:\left(7.84t^2 + 2016t + 169,\!600\right)^{\frac{1}{2}}\)

Differentiate that function with repect to time
. . and substitute \(\displaystyle t = \text{15 minutes} = \text{900 seconds}\)