This gives us the length of each train as 500 meters.

[Ilikebugs]

Two trains of equal length are on parallel tracks. One train is traveling at
40 km/h and the other at 20 km/h. It takes two minutes longer for the trains to
completely pass one another when going in the same direction, than when going
in opposite directions.
Determine the length of each train.

Is there a way to algebraically solve this

 

[MarkFL]

I would convert the speeds to m/min...

\(\displaystyle v\,\frac{\text{km}}{\text{hr}}\cdot\frac{1\text{ hr}}{60\text{ min}}\cdot\frac{1000\text{ m}}{1\text{ km}}=\frac{50}{3}v\,\frac{\text{m}}{\text{min}}\)

And so the speed of the faster train (in m/min) is:

\(\displaystyle v_F=\frac{2000}{3}\)

And the speed of the slower train is:

\(\displaystyle v_S=\frac{1000}{3}\)

In fact, we could write:

\(\displaystyle v_F=2v_S\)

Let's let the length of the trains be $\ell$.

Now, when the trains pass going in the same direction, we have:

\(\displaystyle (v_F-v_S)(t+2)=2\ell\)

or:

\(\displaystyle v_S(t+2)=2\ell\)

And when the trains pass going in the opposite direction, we have:

\(\displaystyle (v_F+v_S)t=2\ell\)

or:

\(\displaystyle 3v_St=2\ell\)

Can you proceed?

 

[Ilikebugs]

t=1 so l=500m?

 

[Greg]

We have

$$3v_St=2L\quad(1)$$

and

$$v_S(t+2)=2L\quad(2)$$

Multiply $(2)$ by 3 and then subtract $(1)$ from the result:

$$3v_S(t+2)=6L\Rightarrow3v_St+6v_S=6L$$

$$6v_S=4L\Rightarrow v_S=\frac{2L}{3}\Rightarrow\frac{1000}{3}=\frac{2L}{3}\implies L=500\text{ m.}$$