We want to find the velocities of two cyclists in the given situation.
The velocity of one cyclist is 17.5 mi/h, and the velocity of the other cyclist is -14.5 mi/h.
The information that we have is that:
Let's say that cyclist 1 starts in the position 0 and has a velocity v₁.
Then the position equation of cyclist 1 is:
p₁(t) = v₁*t
Now cyclist 2 starts at the position 98 miles, and has a velocity v₂, then we can write the position equation of cyclist 2 as:
p₂(t) = v₂*t + 98mi
Assuming that cyclist 2 is the slower one, we can write:
v₂ = -(v₁ - 3mi/h)
Where the negative sign comes because they move in opposite directions.
Then the position equation becomes:
p₂(t) = -(v₁ - 3mi/h)*t + 98mi
Then we know that after 3 hours they are 2 miles apart, then we can write:
p₂(3h) - p₁(3h) = 2mi
Then we need to solve:
-(v₁ - 3mi/h)*3h + 98mi - (v₁*3h) = 2mi
(-6h)*v₁ + 9mi + 98mi = 2mi
(-6h)*v₁ + 107mi = 2mi
(-6h)*v₁ = 2mi - 107mi = -105mi
v₁ = -105mi/(-6h) = 17.5 mi/h
So the velocity of cyclist 1 is 17.5 mi/h
And the velocity of cyclist 2 is:
v₂ = -(17.5 mi/h - 3mi/h) = -14mi/h.
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