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# While on straight road, car X and car Y are traveling at

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While on straight road, car X and car Y are traveling at [#permalink]

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28 Oct 2006, 23:06
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While on straight road, car X and car Y are traveling at different constant rates. If car X is now 1 mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of car Y if 3 minutes ago car x was 1/2 mile ahead of car Y?
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28 Oct 2006, 23:51
If car X is now 1 mile ahead of car Y but 3 min ago was 1/2 mile ahead then car X overtakes car Y with 1/2 miles for 3 min. Then in 6 min car X will be 2 miles ahead of car Y
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29 Oct 2006, 00:51
Hi BG. How do you set-up the equation? Having a tough time with DRT problems lately.
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29 Oct 2006, 06:01
Car X and y are travelling at same speed.
In 3 minutes car x coverd .5 miles now car X is one mile ahed and it needs to cover additional mile

3 ---0.5
x----1
x=3/0.5=6min

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29 Oct 2006, 06:37
Hermione wrote:
Hi BG. How do you set-up the equation? Having a tough time with DRT problems lately.

Hi guys, not sure if this is correct (i think it is ... )

3 Mins ago, Car X was 0.5 miles ahead.
Now, Car X is 1 mile ahead.

:. Car X Travelled 0.5 miles more than Car Y in 3 mins.

Since [Speed of X] - [ Speed of Y] = ([Distance Covered by X] - [Distance Covered by Y] )x time taken by both X and Y to travel.
-> This relationship can be derived from the first level equation Speed = Dist x Time. If you need to, I can put up the derivation here.

:.[Speed of X] - [Speed of Y] = ((1/2)miles / 3mins) = (1/6) miles per min

Now Car X moves (after time t) to a position 2 miles ahead of Car Y.
That means Car X has moved 1 mile ahead of Car Y.

ie. [Distance Covered by X] - [Distance Covered by Y] = 1 mile
-> [Speed of X] x t - [Speed of Y] x t = 1 mile ----> Distance = Spd x time
Since t is the same for both cars, take out common factor;

-> t x ([Speed of X] - [Speed of Y]) = 1 mile
-> t = 1 / (1/6)
-> t = 6 minutes.

ie . 6 minutes from Now, Car X will be 2 miles ahead of Car Y.
29 Oct 2006, 06:37
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