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?
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.
Geometric Solution
