In the first part of the problem, Andrew, traveling at a speed of v1,
travels 4 miles, while Mike, traveling at speed v2, travels (d - 4) miles.
They leave from their respective starting points at the same time, so the
time it takes for them to meet and pass is the same for both. t = d / s
1. t1 = 4/v1 = (d - 4) / v2
Multiply both sides by v1 to eliminate the denominator on the left side,
and multiply both sides by v2 to eliminate the denominator on the right side.
2. (4 / v1) * v1 * v2 = ((d - 4) / v2) * v1 * v2
3. 4v2 = (d - 4) v1
Divide both sides by four to get the value of v2
4. v2 = ((d - 4)v1) / 4
In the second part of the problem, Andrew has reached Simburgh (d)
and turned around, travelling another 2 miles, or (d + 2), while
Mike has reached Kirkton and turned around, travelling another (d - 2) miles,
for a total of d + (d - 2) = (2d - 2) miles.
Again, their times are equal when they meet and pass.
5. t2 = (d + 2) / v1 = (2d - 2) / v2
As in the first part, multiply both sides by v1 to eliminate the denominator
on the left side, and multiply both sides by v2 to eliminate the denominator
on the right side.
6. ((d + 2) / v1) * v1 * v2 = ((2d - 2) / v2) * v1 * v2
7. (d + 2)v2 = (2d - 2)v1
Divide both sides by (d + 2) go get the value of v2
8. v2 = ((2d - 2)v1) / (d + 2)
We have two equations for v2, equation 4 and equation 8. The problem states
that v2 remains the same throughout the journey. Therefore:
((d - 4)v1) / 4 = ((2d - 2)v1) / (d + 2)
Once again, multiply both sides by both denominators.
(((d - 4)v1) / 4) * 4 * (d + 2) = (((2d - 2)v1) / (d + 2)) * 4 * (d + 2)
v1 * (d - 4) * (d + 2) = v1 * (2d - 2) * 4
Divide both sides by v1, eliminating speed from this equation.
(d - 4) * (d + 2) = (2d - 2) * 4
d^2 - 4d + 2d - 8 = 8d - 8
d^2 - 2d - 8 = 8d - 8
Subtract 8d from both sides and add 8 to both sides.
(d^2 - 2d - 8) - 8d + 8 = (8d - 8) + 8d + 8 = 0
d^2 - 10d = 0
Factor out a d on the left side.
d * (d - 10) = 0
One of those factors is equal to 0 (to give a zero answer).
d = 0 doesn't work; we already know the distance is more than 4 miles.
d - 10 = 0
d = 10 <<<<< That's the answer to the first question, how far is it?
We'll substitute that into equation 4 to find v2 in relation to v1.
v2 = ((d - 4)v1) / 4 = ((10 - 4)v1) / 4
v2 = 6v1 / 4 = (6/4)v1
v2 = 1.5 * v1 <<<<<< That's the answer to the second question
No matter what speed you choose for Andrew (v1), Mike's speed is
one-and-a-half times faster.
Let's set Andrew's speed to 6mph and solve equation 1.
t1 = 4/v1 = (d - 4) / v2
t1 = 4mi / 6mph = (10 - 4) / (1.5 * 6mph)
4/6 hr = 6/9 hr
2/3 hr = 2/3 hr
With Andrew travelling at 4 mph, and Mike travelling at 6 mph, it took
both of them 2/3 of an hour to reach a point 4 miles from Kirkton.
