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 Four cars (Posted on 2008-06-09)
Four cars travel at constant speed along a road. Three of them (A, B and C) travel in one direction, and the fourth, D, in the opposite direction.

Car A passes B and C at 9:00h and 10:00h, respectively, and crosses car D at 11:00h.

Car D crosses car B and car C at 13:00h and 15:00h, respectively.

At what time did car B pass car C?

Bonus point: can you solve this geometrically?

 See The Solution Submitted by pcbouhid Rating: 4.2500 (4 votes)

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 Geometric Solution | Comment 3 of 6 |
We can draw a graph of the position over time. Let the horizontal axis represent time, and the vertical axis the position (to make this represent the actual course of the car, the vertical axis might represent the position according to a reference frame moving at constant speed).
We know that all cars travel at constant speed, so all graphs are straight lines.
Draw a horizontal line representing time, and mark the points 9, 10, 11, 13, 15 at according intervals.
Draw an arbitrary line through the point 9. This will represent the course of car A. Get the intersection point P of line A with the vertical line through 11. Draw a line through P and timepoint 15. This represent the progress of car D.
The midpoint M1 of P and point 9 is the point where car A is at time 10. This is at the same position as car C. At timepoint 15, C and D are in the same position. Connecting these gives the course of car C.
Likewise, the midpoint M2 of P and point 15 is where car D and car B meet. Connecting point 9 and M2 gives the trajectory of car B.
When drawing this, it becomes clear how to find the meeting point geometrically. The trajactories of cars B and C are medians of the triangle formed by points 9, 15 and P. The horizontal position is given by (9 + 15 +t(P))/3, where t(P) is the time at point P, or 11.
This gives as a result 11+2/3.

 Posted by Robby Goetschalckx on 2008-06-10 06:39:29

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