The boat

My solution is based on two observations -

Firstly note that the total external force acting on the man-boat system (bouyancy and gravity) is zero, and so there is no acceleration. So the combined system remains at rest. In other words the centre of gravity of the system is stationary.

Secondly note that the total turning moment about the centre of gravity is zero. (otherwise the boat would spin!)

Now suppose that the boat is pointing to the right while the man walks from the back to the front of the boat.

With respect to the stationary centre of gravity the boat moves a distance $x$ to the left, and the man moves a distance of $2.5-x$ to the right.

The movement of the boat to the left increases the anti-clockwise moment by $100x$ while the repositioning of the man increases the clockwise moment by $70(2.5-x)$ .

Since the total moment must remain zero, we have

$100x=70(2.5-x)$

which is easily solved to give

$\boxed{x=1.029}$

Clarification.

When the system is in equilibrium the buoyancy force passes through the centre of gravity of the boat, and so does not contribute to the moments in my solution!

Relevant wiki: Conservation of Momentum

Let $x_T$ be Tim's position and $x_B$ the position of the stern. We choose coordinates so that initially $x_T = x_B = 0$ .

There are no external horizontal forces on the system, so the center of mass $x = (m_Tx_T + m_Bx_B)/(m_T + m_B)$ remains constant, i.e. at all times $m_Tx_T + m_Bx_B = 0.$ When Tim reaches the prow, he is a distance of $L = \SI{2.5}{m}$ from the stern of the boat; i.e. $x_T - x_B = L.$ To eliminate $x_T$ , multiply the bottom equation by $m_T$ and subtract the two equations: $(m_B + m_T)x_B = -m_TL.$ Thus the distance over which the boat moves is $|x_B| = \frac{m_T}{m_B + m_T}L = \frac{70}{100+70}\cdot 2.5=\boxed{1.029}\ \text{m}.$

Considering the conservation of momentum, we have:

$\begin{aligned} m_{Tim} v_{Tim} & = \left(m_{boat}+m_{Tim}\right)v_{boat} & \small \color{#3D99F6} \text{where } m = \text{mass and } v= \text{velocity.} \\ 70 v_{Tim} & = \left(100+70\right)v_{boat} \\ \implies v_{boat} & = \frac 7{17} v_{Tim} & \small \color{#3D99F6} \text{If Tim takes }t \text{ s to walk from the stern to prow.} \\ v_{boat} t & = \frac 7{17} v_{Tim} t \\ s_{boat} & = \frac 7{17} s_{Tim} & \small \color{#3D99F6} \text{where }s = \text{displacement.} \\ & = \frac 7{17} \times 2.5 \\ & \approx \boxed{1.029} \text{ m} \end{aligned}$

The boat and man represent a closed system and therefore conservation of energy applies.

The work done in moving the man in one direction must be balanced by the boat moving in the opposite direction. "Every action has an equal and opposite reaction" Work(j) = Mass(kg) × Distance(m)

Arbitrarily choosing one end of the boat as a reference point and assuming the boat shifts x metres relative to the ground, the man must travel 2.5-x (over the ground) to reach the other end.

And so to conserve energy (mass x distance moved) these two things must balance.

100.x - 70 (2.5-x) = 0

(100+70).x = 70 × 2.5

X = 175/170

If point A is the center of mass of the boat, B the center of mass of the person, and C the center of mass of the boat with the person sitting at one end of the boat, then, when the person moves to the other end of the boat, the boat has to move a distance equal with 2AC in order to keep C from moving. Since AC/1.25 = 70/170, 2AC = 7*2.5/17 = 17.5/17 = 1.029.

The boat’s center of mass (100 kg) is at the origin, and the man’s center of mass (70 kg) is 1.25 m away. The combined system center of mass is therefore (70/170)(1.25) m from the origin. If the man stands at the other end, the system COM is the same distance from the origin, but on the other side. The distance between these points is (70/170)(1.25)(2)= 1.029 m. Since the system COM can’t move without a net external force acting on it, the boat must move this amount.