One bus travels 13 km/h slower than the other.
Let x be the speed of the faster bus.
Then the speed of the slower bus is (x - 10)
The two buses are 860 kilometers apart after 4 hours
Distance = d = 860 km
Time = t = 4 hours
What is the rate (speed) of each bus?
Recall that the relationship between speed, distance, and time is given by
[tex]d=s\cdot t[/tex]Where d is the distance, s is the speed, and t is the time.
The combined speed of the two buses becomes
s = x + x - 10 = 2x - 10
Now let us substitute all the values into the above formula
[tex]\begin{gathered} d=s\cdot t \\ 860=(2x-10)\cdot4 \end{gathered}[/tex]Now let us solve the equation
[tex]\begin{gathered} 860=(2x-10)\cdot4 \\ 860=8x-40 \\ 860+40=8x \\ 900=8x \\ 8x=900 \\ x=\frac{900}{8} \\ x=22.5\: \frac{km}{h} \end{gathered}[/tex]Therefore, the speed of the faster train is 22.5 km/h
The speed of the slower train is (22.5 - 10) = 11.5 km/h
