Step 1: Dynamics and Relative Acceleration

The train stops at $t = u/a_T = 10/2 = 5 \text{ s}$.

The suitcase slides forward relative to the train because the train decelerates faster than the suitcase (due to limited friction). The relative acceleration while the train is moving is:

$$a_{rel} = a_S – a_T = (-\mu g) – (-2) = 2 – 10\mu$$

Since the suitcase slides 12 m, we must determine if the train stops before or after the suitcase stops sliding.

Step 2: Analyzing the Velocity Profile

Relative velocity $v_{rel} = v_S – v_T$.

• Initially, $v_{rel} = 0$.
• As long as the train is moving ($0 < t < 5$), $v_T$ drops rapidly. $v_S$ drops slowly. $v_{rel}$ increases.
• Once the train stops ($t > 5$), $v_T = 0$. The suitcase continues to slide on the stationary floor, slowing down. $v_{rel} = v_S$ decreases until it stops.

This behavior (increase then decrease) confirms statement (c). It also implies the train stops before the suitcase, confirming statement (a).

Step 3: Calculating Coefficient of Friction

Total relative distance slid is the area between the $v_S$ and $v_T$ graphs.

Phase 1 (0 to 5s):

Distance of Train: $S_T = 10(5) – \frac{1}{2}(2)(5)^2 = 25 \text{ m}$.

Distance of Suitcase: $S_{S1} = 10(5) – \frac{1}{2}(\mu g)(5)^2 = 50 – 125\mu$.

Relative displacement in Phase 1: $S_{rel1} = S_{S1} – S_T = 25 – 125\mu$.

At $t=5$, suitcase velocity is $v_{S5} = 10 – 50\mu$.

Phase 2 (t > 5s):

Train is stopped. Suitcase stops from velocity $v_{S5}$ with deceleration $\mu g$.

$$S_{rel2} = \frac{v_{S5}^2}{2\mu g} = \frac{(10 – 50\mu)^2}{20\mu}$$

Total Distance:

$$S_{total} = S_{rel1} + S_{rel2} = 12$$ $$25 – 125\mu + \frac{(10 – 50\mu)^2}{20\mu} = 12$$

Let $x = 10\mu$. The equation becomes:

$$25 – 12.5x + \frac{(10-5x)^2}{2x} = 12$$ $$50x – 25x^2 + 100 – 100x + 25x^2 = 24x$$ $$100 – 50x = 24x \implies 74x = 100 \implies x \approx 1.35$$ $$\mu = \frac{x}{10} = 0.135$$

This confirms statement (d) is correct.