Thank you Barbara and Colin prompt responses.
I think it really depends on how you view the equation or how you derive it. There are two very distinctive approaches.
Just using definition of a= dv/dt and v=ds/dt
= dv/ds * ds/dt
a= v * dv/ds
a ds= v dv
then integrate both sides by putting when s=0, v=u and when s=s v=vwe will have as = 1/2 v^2 -1/2 u^2
Work = force * distance
dW = F* ds (in here, the * means the vector dot product)
input work = output work. (conservation of energy)
Gravitational energy = work done to the particle with mass m.
(mg) * ds = F *ds
= ma *ds
= m (dv/dt) * ds
= m dv* (ds/dt)
= m v * dv
The left hand side is just mgh.
Canceled the m from both sides, this will also result in exactly the same equation.
Please note in the method 2, mg and s are in vector form. Although the vector acceleration a in method 1 is not clear for the track, I still can use g as the acceleration and h as displacement. But as Barbara mentioned, it really doesn’t matter as g will always parallel to h.
The first approach used the definition of a and v. The result from method 1 is totally just a mathematical results of manipulation of symbols and chain rule.
Method 2 showed the conservation of energy that KE = Force* distance in different form.
That is why I puzzled about the examiner report which regarded students using v^2=u^2 +2as is wrong. In my opinion, this kinematic equation or KE=PE are just the same.