dv/dt = 10
v = INTEGRATE (dv/dt) dt
v = INTEGRATE 10 dt
v = 10t + c

We know that the bungee jumper starts out at rest, so v = 0 when t = 0

0 = 10 (0) + c
c = 0

So, v = 10t

Her velocity at X (t = 4)
= 10 (4)
= 40 m/s

Integrating v, we get
s = INTEGRATE v dt
s = 5t² + d (use a different letter for the constant since c has already been used)

Of course the bungee jumper starts out from O (which we take to be the 0 m mark), so s = 0 when t = 0

0 = 5 (0)² + d
d = 0

s = 5t²

When t = 4, s = 5 (4)² = 80 m

This indicates that the bungee jumper is 80 m from O (since O is represented by s = 0), so the distance OX is 80 m.

"On reaching X, she then slows so that her velocity, V m/s, T seconds after reaching X, is such that dV/dT = 10 - kT, where k is a constant"

Basically, we now start the question all over, but using X as the starting reference point rather than O. So we ignore what has happened earlier (except for the values related to X).

In particular, V = 40 when T = 0.

dV/dT = 10 - kT

Integrating,
V = INTEGRATE (dV/dT) dT
V = INTEGRATE (10 - kT) dT
V = 10t - 0.5kT² + f (I cannot use e as a constant to avoid confusion with the exponential constant e)

Since V = 40 when T = 0.
40 = 10 (0) - 0.5k (0)² + f
40 = f

V = 10t - 0.5kT² + 40

Since she first comes to rest at a point Y when T = 3, V = 0 when T = 3, so

0 = 10 (3) - 0.5k (3)² + 40
0 = 30 - 4.5k + 40
4.5k = 70

Multiplying both sides by 2,

9k = 140
k = 140/9