SOLUTION: Person on 3rd base walks at a rate of 6' per sec., person on 1st base jogs at a rate of 10' per sec. How long will it take them to be 270' apart? My teacher says its not a time,

Question 158798: Person on 3rd base walks at a rate of 6' per sec., person on 1st base jogs at a rate of 10' per sec. How long will it take them to be 270' apart? My teacher says its not a time, distance, rate problem but a triangle problem. I'm confused.
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Well, it's both.
Here's where they are at first.

If we set up a coordinate system at first base then the person at first base is at (0,0).
The x-axis runs from first to second.
The y-axis runs from first to home plate.
The person at third base is at (90,90).
The position of the person at first changes with t by
(10t,0)
and the position of the person on third base changes with t by
(90-6t,90).
The x distance is one leg of the triangle, determined with rate and time.
The y distance, fixed at 90 feet, is another leg of the triangle.
The distance is the hypotenuse of the triangle.
The distance between the two, using the distance formula, is

Find t when D=270.

Use the quadratic formula,

Only use the positive value,

Let's check the answer.
After 21.5 seconds, the person on 1st will be at
(10t,0)=(10(21.5),0)=(215,0)
and the person on third will be at
(90-6t,90)=(90-6(21.5),90)=(-39,60)
The distance is then

Good answer.
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