Question
A car has an initial velocity of 12 m/s and is brought to rest over a distance of 45 m. The acceleration of the car is
NDA
General Ability
General Science
2025
NDA 2025 (Phase 1)
NDA General Ability Question (2025) — Solution
Options
- A. +1.6 m/s^2
- B. +3.2 m/s^2
- C. -1.6 m/s^2
- D. -0.8 m/s^2
Answer
C. -1.6 m/s^2
Step-by-step solution
Given initial velocity u = 12 m/s. Final velocity v = 0 m/s. Distance covered s = 45 m. Using the third equation of motion: v^2 = u^2 + 2as Substituting the given values: 0 = (12)^2 + 2a(45) 0 = 144 + 90a 90a = -144 a = - 144 90 a = -1.6 m/s^2 Answer: -1.6 m/s^2
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