|Feb7-09, 10:56 PM||#1|
Question why is there a negative sign
A particle moves along the c axis. its position is given by the equation x = 2+3t-4t^2, with x in meters and t in seconds. Determine (a) its position when it changes direction and )b) its velocity when it returns to the position it had at t = 0.
i understand how to get (a) but for (b) im confused why in the solution they put -x = -2-3t+4t^2 why is the negative there, i understand there is a change in direction at 0, im confused why the negative sign is there can you give me a clear and understanding reasoning. thank you.
|Feb8-09, 01:23 AM||#2|
I have no clue. They just randomly decided to flip the signs on everything, it looks like =/
|Similar Threads for: Question why is there a negative sign|
|Why is the sign in the formula Δy=VoyT - (1/2)atē negative in this problem||Introductory Physics Homework||2|
|Gravitational Force(Why the negative sign?)||Introductory Physics Homework||5|
|sign question||Set Theory, Logic, Probability, Statistics||3|
|passive sign convention (negative watts, and negative current confusion)||Advanced Physics Homework||7|
|Does this negative sign even matter in sin(x)?||Calculus & Beyond Homework||5|