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?

Question

happykiddo · Answer

Could someone explain to me what a "turning point" is in relation to this problem?

happykiddo · Answer

The answer is 5, if that helps any.

emmatassone · Answer

the function is:$$x(t)=t ^{2}+t-2$$
right?

happykiddo · Answer

So sorry  the actual function is            x=(-2t^2+0t+5)
not the one stated above.

happykiddo · Answer

not the one in the question above. Function different overall question the same.

happykiddo · Answer

sorry for the confusion.

emmatassone · Answer

the turning point is a point at which the derivative changes sign

emmatassone · Answer

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.

happykiddo · Answer

I understand what I must do now. Thank you very much!  : )

emmatassone · Answer

no problem