If the particle’s position on the x-axis is now given by the function x=(1t^2+1t+-2)m, where t is in s. To the nearest 0.1 m, what is the turning point (where the particle reverses direction of motion) of the particle's position?
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Could someone explain to me what a "turning point" is in relation to this problem?
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So sorry the actual function is x=(-2t^2+0t+5)
not the one stated above.
not the one in the question above. Function different overall question the same.
sorry for the confusion.
the turning point is a point at which the derivative changes sign
so you should derivate the function and see when the derivative becomes zero. And when you find that turning point then you will have to evaluate in your position function to know at what position was the particle when it reverse his direction of motion.
I understand what I must do now. Thank you very much! : )