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Question :

Two trains start from the same station at the same time. One goes due north at 30 miles per hour, and the other due north-east at 20 miles per hour. At what rate is the distance between them increasing after $1/4$ hours?


I need to find the speed that these two trains $T_1$ and $T_2$ are moving away from each other at time $t = \frac{1}{4}$ hours, given:

enter image description here

From that image then, in order to find the rate that line $DB$ changes, I need to find the rate that $AB$ changes, and the rate that $AD$ changes.

For $AB$, this is only dependent on $T_1$, so I have :

$$ \frac{dx}{dt} = \Delta s \times \frac{1}{\sqrt{2}} $$

Where $\Delta s = 20 mph$ and at 15 minutes this would be $\Delta s = 4$, giving

$$ \frac{dx}{dt} = 4 $$

For $\frac{dy}{dx}$ I have to find the difference between the change in $y$ for $T_2$ and the change in $y$ for $T_1$.

For $T_2$ I have

How x, y change with respect to $T_2$

$$ \frac{dy}{dx} = \Delta s \times \frac{1}{\sqrt{2}} $$

For $T_1$ I have, at $t = \frac{1}{4}$ hours

$$ \frac{dy}{dx} = 7.5 $$

Then the overall change in $y$ would be

$$ \frac{dy}{dx} = 7.5 - 4 \times \frac{1}{\sqrt{2}} = \frac{15 - 4\sqrt{2}}{2} \approx 4.67 $$

From this I have a triangle as

enter image description here

Which gives the change of $DB$ as

$$ \frac{d(DB)}{dt} = \sqrt{4^2 + 4.67^2} \approx 6.149 $$


I think the above makes sense, but I'm not confident about the approach.

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closed as off-topic by John Rennie, Kyle Kanos, ZeroTheHero, Yashas, Jon Custer Apr 30 '17 at 18:20

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I am not going to answer it for you as it is better if you work it out yourself but here is a useful hint.

Convert speeds to distance per MINUTE. Get some graph paper, plot the positions every minute, observe how the distance between T1 & T2 represented by a line on the graph paper moves as the minutes increase.

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