Step 1: Accelerations

Plank: The plank stops from $2.0 \text{ m/s}$ in $0.5 \text{ s}$.

$$a_P = \frac{v – u}{t} = \frac{0 – 2.0}{0.5} = -4.0 \text{ m/s}^2$$

Block: The only horizontal force on the block is friction from the plank. Since the plank slows down rapidly, the block will slip forward relative to the plank. The maximum deceleration provided by friction is:

$$a_B = -\mu g = -0.10 \times 10 = -1.0 \text{ m/s}^2$$

Step 2: Relative Motion during Plank Deceleration ($0 \le t \le 0.5$)

Relative acceleration of the block with respect to the plank:

$$a_{rel} = a_B – a_P = (-1.0) – (-4.0) = +3.0 \text{ m/s}^2$$

Relative distance covered in $t = 0.5 \text{ s}$:

$$s_1 = u_{rel}t + \frac{1}{2}a_{rel}t^2 = 0 + \frac{1}{2}(3.0)(0.5)^2 = 0.375 \text{ m}$$

At $t = 0.5 \text{ s}$, the plank stops ($v_P = 0$). The block’s velocity is:

$$v_B = u + a_B t = 2.0 + (-1.0)(0.5) = 1.5 \text{ m/s}$$

Step 3: Motion after Plank Stops ($t > 0.5$)

Now the plank is stationary. The block moves at $1.5 \text{ m/s}$ on the stationary plank. It continues to decelerate at $a_B = -1.0 \text{ m/s}^2$ until it stops.

Distance covered in this phase:

$$v^2 – u^2 = 2 a s_2$$ $$0 – (1.5)^2 = 2(-1.0) s_2$$ $$2.25 = 2 s_2 \implies s_2 = 1.125 \text{ m}$$

Step 4: Total Distance

The total distance slid by the block on the plank is the sum of the distances from both phases:

$$S_{total} = s_1 + s_2 = 0.375 + 1.125 = 1.5 \text{ m}$$