We can model the problem as follows: on a number line we put train A, with positive speed
, at the origin x=0.
At x=480 we put train B, with negative speed
.
So, the equations for the positions of the two trains are
![\begin{cases}x_A = v_At\\x_B=480-v_Bt\end{cases}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/k3b1j6bamr5y2nph3c1tabled4v90mn1bz.png)
We know that
, so we can rewrite the first equation:
![\begin{cases}x_A = (v_B+5)t\\x_B=480-v_Bt\end{cases}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2k3bvbrstijwtu74pp24t62x5nv5j93osi.png)
The two trains meet when they are at the same position:
![x_A=x_B \iff (v_B+5)t=480-v_Bt](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xdr4is8w7fenrcpfx8s912grfxaqh455hi.png)
We know that this happens after 3 hours, i.e. when t=3:
![3(v_B+5)=480-3v_B \iff 3v_B+15=480-3v_B \iff 6v_B = 465 \iff v_B=77.5](https://img.qammunity.org/2020/formulas/mathematics/middle-school/a7t4pnmlawy6nzl2mfzy9jw60otwgi40vx.png)
And since train A was 5 mph faster, we have
![v_A=77.5+5=82.5](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ia56r2k9op5unaffg00l7buqd1866vrw6i.png)