Since s(t) is the position, we can take the derivative to get the equation for the velocity.
i did that and got this but the answer is not correct:s(t)=2t 3 −24t 2 +90t v(t)=s ′ (t)=6t 2 −48t+90=6(t−3)(t−5) a(t)=s ′′ (t)=12t−48=12(t−4)
(A) forward 0<t<3 or t>5, backward 3<t<5(B) speeding up t>4, slowing down 0<t<4
Actually, I need to correct that.
Forward is [0,3) union (5,infinity)slowing down is [0,4) and speeding up is (4,infinity)