Aparticle's position is given by x = 4.0t2 - 32t + 36, with x in meters and time t in seconds. at what position does the particle come to a momentary stop, to reverse its motion?

Firstly, we must find the equation the speed, it can be obtained by the derivative of x =x = 4.0t2 - 32t + 36, it means v= 8.0t-32. the particule stops means v= 0, and 0= 8.0t-32, which implies 8t=32, so t=4s, and then x(t=4) = 4.0(4)^2 - 32(4) + 36= - 28m
so x = - 28 m

So what is the question