what would be the equation be for change in distance, using x for velocity initial, z for time, y for acceleration, and using units of meters and seconds? (as simplified as possible) I have a quiz tommorow, and I know the actual formula, and how to use it to find an answer with actual numbers, but the quiz is with variables instead
Thanks for the help!
2007-03-05 10:54:24 · 3 answers · asked by makelovenothate 2 in Science & Mathematics ➔ Physics
Change in distance = .5*acceleration*(time^2) + initial velocity*time
With your indicated variables...
Change in distance = .5y(z^2) + xz
z^2= z*z... Just in case that notation was confusing.
2007-03-05 11:00:32 · answer #1 · answered by Frigidaire 1 · 0⤊ 0⤋
(z(x+yz)+zx)/2=distance travelled
The basic formula is to find the change in speed, add it to initial speed, then multiply by time. Then just find the initial speed and multiply that by time. then you need to find the average of those two.
I'll simplify the formula a bit more
zx+0.5yz^2
2007-03-05 19:01:29 · answer #2 · answered by MLBfreek35 5 · 0⤊ 0⤋
distance = xz + 1/2yz^2
2007-03-05 19:00:07 · answer #3 · answered by physicist 4 · 0⤊ 0⤋