Math destroyer problem

[jh311]

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I have a question on how to set this question up, i thought there were 2 ways of doing it, but only one of the ways works, please help me understand why only this way works:

Question 31 test#6 2010MathDestroyer:

Car A sets out on a journey at a stead speed of 45MPH. Two hours later Car B sets out from the same starting point along the same route, traveling at a steady rate of 80mph. Approximately how long will it take Car B to catch up with Car A?

So, since their distances are the same when the second car caches up you can set it up like this:

(VelocitycarA)(TimeCarA) = (velocityCarB)(timeCarB)

since car has been on the road 2 hours longer than car B, you can call his time t+2, and carB's time t

45(t+2) = 80t
45t+90 = 80t
90=35t
t= 90/35 = about 2.5hrs

my question is: why can't you call the time of carB = t-2, since it's been on the road 2hrs less than carA????
doing that would make it this:
45t = 80(t-2)
45t = 80t-160
160=35t
t = 160/35 = about 4.5hrs


I don't understand why you can only set it up the first way and not the second way....please help!! Thanks

 

[AJS11]

You CAN set it up the way you did it, but remember you defined "t" as Car A's time. So plug in your value of t into your equation for car B.

t-2 = Car B

(4.5)-2 = 2.5 which is the answer.

 

[jh311]

oh I See!!! That helps a lot, thanks. I just forgot what i had set equal to what and what the question was really asking.