Answer:
77.5 mph and 82.5 mph
Step-by-step explanation:
Let's say the speeds of the trains are x and x + 5.
The combined distance traveled by the trains is 480 miles.
480 = x × 3 + (x + 5) × 3
480 = 3x + 3x + 15
480 = 6x + 15
465 = 6x
x = 77.5
The speeds of the trains are 77.5 mph and 82.5 mph.
We can model the problem as follows: on a number line we put train A, with positive speed [tex]v_A[/tex], at the origin x=0.
At x=480 we put train B, with negative speed [tex]v_B[/tex].
So, the equations for the positions of the two trains are
[tex]\begin{cases}x_A = v_At\\x_B=480-v_Bt\end{cases}[/tex]
We know that [tex]v_A=v_B+5[/tex], so we can rewrite the first equation:
[tex]\begin{cases}x_A = (v_B+5)t\\x_B=480-v_Bt\end{cases}[/tex]
The two trains meet when they are at the same position:
[tex]x_A=x_B \iff (v_B+5)t=480-v_Bt[/tex]
We know that this happens after 3 hours, i.e. when t=3:
[tex]3(v_B+5)=480-3v_B \iff 3v_B+15=480-3v_B \iff 6v_B = 465 \iff v_B=77.5[/tex]
And since train A was 5 mph faster, we have
[tex]v_A=77.5+5=82.5[/tex]
