camboguy
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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) I am 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, I am confused why the negative sign is there can you give me a clear and understanding reasoning. thank you.
i understand how to get (a) but for (b) I am 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, I am confused why the negative sign is there can you give me a clear and understanding reasoning. thank you.