Applied Maths - Mechanics 44

I try to solve the problem only mathematically.

Let x be the displacement of the mass from the original position.
At time t, the other end of the string has moved a distance of ut.
Total extension of the string = ut – x
md2x/dt2= (mg/a)(ut – x)
d2x/dt2 + gx/a = gut/a … (1)
General solution is x = Asin√(g/a)t + Bcos√(g/a)t
Let the particular solution be Ct2 + Dt + E
Put back into (1) 2C + (g/a)(Ct2 + Dt + E) = gut/a
Therefore C = 0, E = 0 and D = u
The combined solution is x = Asin√(g/a)t + Bcos√(g/a)t + ut
At time t = 0, x = B = 0
dx/dt = = A√(g/a)cos√(g/a)t – B√(g/a)sin√(g/a)t + u
At time = 0, dx/dt = 0 => A√(g/a) + u = 0
A = – u√(a/g)
Therefore the complete solution is x= – u√(a/g)sin√(g/a)t + ut … (2)
while dx/dt = – ucos√(g/a)t + u … (3)
The string is stretched as far as ut >= x, so (2) =>
u√(a/g)sin√(g/a)t >= 0
This is true as far as √(g/a)t <= π
So the time for the string to remain stretched is π√(a/g)
The distance traveled so far by (2) when t = π√(a/g) is πu√(a/g)
The velocity of the mass at such point is 2u from equation (3)
In other words, after time t, the particle is faster than the other end, and the string will never become stretched again.
In order for the mass to catch up, it needs another time t1 such that (2u – u)t1 = a
t1 = a/u
Additional distance traveled by the mass is 2ut1 = 2a
The total distance traveled by the particle = πu√(a/g) + 2a

I copy the question from my book. No more additional information is provided.

Would you please provide the acceleration of the driving force
that pull the string?

Otherwise, it is impossible to answer the 2 questions.