Rates of relation word problem

[Guest]

Two cyclists depart at the same time from a starting point along routes making an angle of pi/3 radians with each other. The first is travelling at 25 km/h, while the second is moving at 20km/h. How fast are the two cyclists moving apart after 2h?

I don't get this question >.< I have absolutly no idea why u need the angle too

 

[pka]

Use the angle between the riders in the law of cosines.
If x is the distance between rider a and rider b then:
\(\displaystyle \L
\begin{array}{l}
x^2 = a^2 + b^2 - 2ab\cos (\pi /3) \\
x^2 = a^2 + b^2 - ab \\
2x\frac{{dx}}{{dt}} = 2a\frac{{da}}{{dt}} + 2b\frac{{db}}{{dt}} - \left( {\frac{{da}}{{dt}}b + a\frac{{db}}{{dt}}} \right) \\
2x\frac{{dx}}{{dt}} = \left( {2a - b} \right)\frac{{da}}{{dt}} + \left( {2b - a} \right)\frac{{db}}{{dt}} \\
\end{array}\).

After two hours a=50, b=40 and x=60.
Now you solve this.

 

[soroban]

Hello, bittersweet!

Two cyclists depart at the same time from a starting point along routes at an angle of 60^owith each other.

The first is travelling at 25 km/h, while the second is moving at 20km/h.

How fast are the two cyclists moving apart after 2h?

I have absolutly no idea why u need the angle too

Suppose the angle was 0^o.
Then they are going in the same direction down the same road.
$\displaystyle \;\;$They are separating at: $\displaystyle 25\,-\,20\:=\:5$ km/h.

Suppose the angle was 180^o.
They they are going in opposite directions down the road.
$\displaystyle \;\;$They are separating at: $\displaystyle 25\,+\,20\:=\:45$ km/h.

Can you see that the angle makes a difference?

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pka's suggestion is the best: the Law of Cosines.

I would apply it differently. $\displaystyle \;$(Is anyone surprised?)

                B
                *
              /  \
         20t/     \ x
          /        \
        / 60°       \
      * - - - - - - -*
      O     25t      A

The cyclists (#1 and #2) start at point $\displaystyle O$ at the same time.

In 2 hours, #1 has gone $\displaystyle 25t$ km to point $\displaystyle A$.
In 2 hours, #2 has gone $\displaystyle 20t$ km to point $\displaystyle B$.

Let $\displaystyle x$ = distance beween $\displaystyle A$ and $\displaystyle B$.

Law of Cosines: \(\displaystyle \:x^2\:=\25t)^2\,+\,(20t)^2\,-\,2(25t)(20t)\cos60^o\;\;\Rightarrow\;\;x^2\:=\:625t^2\,+\,400t^2\,-\,1000t^2(0.5)\)

Then: $\displaystyle \:x^2\;=\;525t^2\;\;\Rightarrow\;\;x\:=\:5\sqrt{21}t$


Differentiate with respect to time: $\displaystyle \:\frac{dx}{dt}\:=\;5\sqrt{21}$

The rate of change is constant.

Therefore, they are always separating at $\displaystyle 5\sqrt{21}\,\approx\,23$ km/h.