physics question help please

hi, please could I have some help with question 2c of this paper? Shouldn't the graph start at 0.7 then decrease to 0 because it starts in the middle then moves down?
paper: https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FPhysics%2FA-level%2FPast-Papers%2FCAIE%2FPaper-2%2FQP%2FNovember%202021%20(v2)%20QP.pdf
thanks!

You are correct saying in should start at 0.7*10^-2 m then decrease to 0.
For the shape of the curve:
F = EQ and F = ma.
Combining these, EQ = ma --> a = EQ / m
From suvat, s = ut + 1/2 * at² , u = 0 as the particle starts from rest
--> s = 1/2 * at² which rearranges to a = 2s / t²
Putting the two acceleration equations together, a = EQ / m = 2s / t²
Rearrange to find s, s = QE / 2m * t2
As x is getting smaller, but s is getting larger when t increases, x = -s
--> x = - QE / 2m * t²
The shape of the graph is a negative parabola with increases magnitude of gradient

but this is what the ms says: line from (0, 0.7 × 10–2) to a non-zero point on the t-axis M1 magnitude of gradient of line increases

For the first marking point: the particle does not drop instantaneously so will take actual time to drop. The line will intersect the t-axis somewhere between 0 and infinity.
For the second marking point, the curve has the equation in the form x = - k t^2 which I showed in my previous post. The gradient is increasingly negative as the tangent gets steeper as t increases. (dy/dx = - 2kt) Therefore the magnitude of the gradient increases as t increases

Ohhh ok, but why does it not touch the t-axis, does x become 0 when it reaches the bottom plate?

x does become 0 when the particle touches the bottom plate so the line stops when it intersects the t-axis