I'm thinking there is not one, but two ways, to do a linear combination of two vectors if I include time. $$ $$ For example: $$ $$ If a point $p$ has an i-ward velocity of 1 meter per second and a j-ward velocity of 1 meter per second, then it will move from (0,0) to (1,1) in one second. $$ $$ If on the other hand, it has an i-ward velocity of 2 meters per second for HALF a second, then replaced by a j-ward velocity of 2 meters per second for HALF a second, then it will ALSO move from (0,0) to (1,1) in one second. $$ $$

One of these is: $$ (\vec i + \vec j ) . \delta t $$

$$$$ and the other is $$ (\vec 2i + \vec 2j ) . \frac{\delta t}{2} $$

In the first case, the i and j components act simultaneously, in the second case they act sequentially.

So, given that $p$ goes from (0,0) to (1,1) in unit time in both cases, how could one know by which path it had done so if all one had to go on was the starting and ending points of $p$ and its starting and ending times?

  • $\begingroup$ I feel like you asked this same question (with not as much detail) earlier today? If it was closed (or something like that), the correct thing to do is to edit your original post and request that it be reopened. $\endgroup$
    – march
    Feb 23 at 22:18
  • $\begingroup$ You have to know the speed of the particle $v = \sqrt{ v_x^2 + v_y^2}$. This will let you decompose the motion into a magnitude (speed) and a direction. $\endgroup$ Feb 23 at 22:38

1 Answer 1


If you only consider the start and finish, you cannot know the path. Consider this situation. In one case, you walk in a large circle, speeding up to a maximum speed and then slowing back to zero speed as you reach your original location. In another case, you stand still for the same amount of time. Start and finish are not enough to determine what happened during the event. We can sometimes fit a predetermined model to the event, such as projectile motion or straight-line motion. Without a defined pattern or path from start to finish, beginning and end cannot define the middle.


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