A high-speed maglev train accelerates uniformly from rest to $108\text{ km/h}$ over a distance of $450\text{ m}$. Assuming constant acceleration, the time taken (in seconds) to cover this distance is:

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### 1. Strategy & Theoretical Principles Convert final speed to SI units: $$v = 108\text{ km/h} = 108 \times \frac{5}{18} = 6 \times 5 = 30\text{ m/s}$$ For uniform acceleration from rest ($u = 0$) over distance $s$: $$s = \frac{u + v}{2} t = \frac{v}{2} t$$ ### 2. Analytical Formulation & Calculation $$450 = \frac{30}{2} t = 15 t$$ $$t = \frac{450}{15} = 30\text{ seconds}$$ ### 3. Distractor Forensics - Candidate forgetting to convert $\text{km/h}$ to $\text{m/s}$ gets $450 / 54 = 8.33\text{ s}$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Always multiply $\text{km/h}$ by $5/18$ to convert to $\text{m/s}$. - **Standard Reference:** GATE General Aptitude Official Syllabus.

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