Question 1139968
If we count time from the moment the buses are approaching each other but still 13 miles apart, the distance {{{D}}} (in miles) between the buses as a function of time {{{t}}} (in hours) is
{{{d=f(t)=sqrt((13-110t)^2+5^2)}}}
 
{{{drawing(500,175,-10,10,-1,6,
green(arrow(0,0,12,0)),green(arrow(0,0,-12,0)),
green(arrow(0,5,12,5)),green(arrow(0,5,-12,5)),
red(line(-6.5,0,6.5,0)),red(line(-6.5,0,-6.5,5)),
red(line(6.5,0,-6.5,5)),circle(6.5,0,0.2),
arrow(6.5,0,9,0),circle(-6.5,5,0.2),
red(rectangle(-6.5,0,-6,0.5)),
arrow(-6.5,5,-9,5),locate(-1.5,-0,red(13-110t)),
locate(-6.4,2.8,red(5)),locate(-1.5,3,red(D)),
locate(6.2,-0.2,bus),locate(-7,5.8,bus)
)}}}
As the buses are passing each other, their distance is about 5 miles, and not changing very sharply. When they are very far from each other their distance is changing by almost 110mph. At a time {{{t}}} , the rate of change (in mph) for the distance between the buses is the value of the derivative of the function {{{f(t)}}} above.
That derivative, {{{df/dt}}} or {{{"f'"}}} is the function
{{{"f ' ( t )"=-110(13-110t)/sqrt((13-110t)^2+5^2)}}} , and its value at {{{t=0}}} is
{{{"f ' ( 0 )"=-110(13)/sqrt(13^2+5^2)=-1430/sqrt(194)}}} .
It is a negative value, because the distance is decreasing, and in absolute value is approximately {{{102.66}}} .
When the buses are at 13 miles from each other, approaching each other, the distance between them is decreasing at {{{highlight(approximately 102.66mph)}}} .