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November 27, 2014

November 27, 2014

Posted by **Al** on Sunday, May 27, 2012 at 8:41pm.

- Algebra -
**Reiny**, Sunday, May 27, 2012 at 9:20pmDepends how far back the faster car was.

Not enough data.

- Algebra -
**Al**, Sunday, May 27, 2012 at 9:21pmLeaves 2 minutes after the car at 80 mph passes it

- Algebra -
**Al**, Sunday, May 27, 2012 at 9:23pmThe car traveling 100 mph states from a complete stop

- Algebra -
**Reiny**, Sunday, May 27, 2012 at 10:28pmKey point:

When the faster catches the slower , they both went the same distance

let that distance be x miles

time for faster car = x/100

time for slower car = x/80

x/80 - x/100 = 2/60

x/400 = 1/30

30x = 400

x = 40/3 miles

so time for faster car to go the 40/3 miles to catch the slower car

= (40/3) / 100 = .13333.. hr

=**8 minutes**

check:

time for slower car

= (40/3) / 80 = .166666 hrs

= 10 minutes, for a difference of 2 minutes

- Algebra -
**Al**, Sunday, May 27, 2012 at 10:38pmThanks!

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