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I need help with math motion problems.
Ex. Two cars left a shop. The first car travels 55 mph and left at 9:00 am. The other car left one hour later at 75 mph. At what time did the second car catch up with the first?

Verified Answer :

AL2006

The whole story begins at 9:00 AM, so let's make up a quantity called ' T ',
and that'll be the number of hours after 9:00 AM.  When we find out what ' T ' is,
we'll just count off that many hours after 9:00 AM and we'll have the answer.

-- The first car started out at 9:00 AM, and drove until the other one caught up
with him.  So the first car drove for ' T ' hours.

The first car drove at 55 mph, so he covered ' 55T ' miles.

-- The second car started out 1 hour later, so he only drove for (T - 1) hours.

The second car drove at 75 mph, so he covered ' 75(T - 1) ' miles.

But they both left from the same shop, and they both met at the same place.
So they both traveled the same distance.

(Miles of Car-#1) = (miles of Car-#2)

55 T = 75 (T - 1)

Eliminate the parentheses on the right side"

55 T = 75 T - 75

Add 75 to each side:

55 T + 75 = 75 T

Subtract 55 T from each side:

75 = 20 T

Divide each side by 20 :

75/20 = T

3.75 = T

There you have it.  They met 3.75 hours after 9:00 AM.

9:00 AM + 3.75 hours = 12:45 PM . . . just in time to stop for lunch together.

Also by the way ...
when the 2nd car caught up, they were 206.25 miles from the shop.

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