NDA2025General AbilityGeneral ScienceActual
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
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
Correct 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