One train travels north at $140$ mph towards Traveler's Town, while a second train travels west at $150$ mph away from Traveler's Town.

Question

One train travels north at $140$ mph towards Traveler's Town, while a second train travels west at $150$ mph away from Traveler's Town. At time $t=0$, the first train is $70$ miles south and the second train is $90$ miles west. Find the rate at which the distance between the trains is changing at time $t=30$ minutes

I have thought to do the following:

I made a drawing that illustrated the situation and gave me a right triangle. Suppose that the distance of the train one to the origin is $x$ and that the distance of the train two to the origin is $y$, then in the time $t=0$ we have that $x=-70$ and $y=-90$ and the distance $d$ between the two is $d=114,02$. I have thought to use the pythagoras theorem and implicit derivation to get to that $x^2+y^2=d^2$ and $2x\frac{dx}{dt}+2y\frac{dy}{dt}=2d\frac{dd}{dt}$ but I do not know what else to do, I know that $\frac{dx}{dt}=140$ and $\frac{dy}{dt}=150$ but here I am stuck, I would appreciate any collaboration, thank you very much.

Answer 1

Place the town in the origin. Now let the first axis point due east, and the second axis due north. Now the trains can be parametrized by: \begin{align} \vec{r_1}(t) &= \langle 0, -70+140t\rangle\\ \vec{r_2}(t) &= \langle -90-150t,0\rangle . \end{align}

Hence the distance between the trains is: \begin{align} d(t) = \sqrt{(\vec{r_2}(t)-\vec{r_1}(t))^2}= 10 \sqrt{421t^2+74t+130}. \end{align} Now all we need to find is $d'(1/2)$, which after some calculations is 150 mph.

Answer 2

The first train is $70$ miles south and moving north at $140$ mph, so after $t$ hours it will be $70-140t$ miles from the Traveler's Town.

The second train is $90$ miles west and moving west at $150$ mph, so after $t$ hours it will be $90+150t$ miles from the Traveler's Town.

The distance between the two trains is (by Pythagoras): $$d=\sqrt{(70-140t)^2+(90+150t)^2}.$$ The distance is changing at time $t=0.5$ hours at the rate: $$d'(0.5)=\frac{-2(70-140\cdot 0.5)\cdot (-140)+2(90+150\cdot 0.5)\cdot 150}{2\sqrt{(70-140\cdot 0.5)^2+(90+150\cdot 0.5)^2}}=150 \ \text{mph}.$$