Two cars leave a shop at different times.
- The first car travels at 55 mph and leaves at 9:00 am.
- The second car leaves one hour later at 75 mph.
At what time does the second car catch up with the first?
Answer (2)
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.
The second car catches up with the first car at 12:45 PM. The first car travels for 3.75 hours from its departure at 9:00 AM to the point of catching up. This occurs after the second car begins its journey one hour later at a higher speed.
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