View Full Version : Final displacement
lillib123
Sep 13, 2009, 11:36 AM
When calculating for final displacement, what is the difference between these two equations?
1. x(f) = x(o) + V(o)t + at2 / 2
2. x(f) = x(o) + (v+v(o)) / 2
x(f) = final displacement, x(o) = original displacement, v(o) = original velocity.
It's my fourth day taking AP physics, and I need help.. thanks!
crigby
Sep 13, 2009, 12:31 PM
Hi,
Off the bat I would say they are probably the same with different varaibles for solution (you did not define a or t but I assume them to be acceleration and time.) The second does assume constant velocity in its calculation.
Peace,
Clarke
Unknown008
Sep 14, 2009, 07:53 AM
The second one cannot be used to find the displacement. You see, the sum of the initial and final velocities when divided by two cannot be added to a displacement.
However, in the first one, you have a velocity times a time, giving a displacement, and an acceleration times a time to the square. Acceleration times time give velocity, then times time again give displacement.
Thus, the second equation is one that does not exist.
Hope it helped! :)
lillib123
Sep 14, 2009, 06:00 PM
[QUOTE=Unknown008;1977830] You see, the sum of the initial and final velocities when divided by two cannot be added to a displacement.
QUOTE]
Why not?
Unknown008
Sep 15, 2009, 06:56 AM
That just isn't possible to add a distance with a speed! If I tell you add a temperature with another temperature, that's possible. But if I tell you to add a temperature with a volume, that's not possible.
Speed and distance are two different things, and cannot be added together.
Taking the two velocities added up, and divided by two will give an 'average' of the velocities, which is still a velocity. Adding distance with that cannot be done.
Hope it cleared some misunderstandings! :)
lillib123
Sep 15, 2009, 12:27 PM
Thanks!