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
- 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}
Explanation
Using the equation of motion v^2 - u^2 = 2as. Here, v = 0, u = 12 m/s, and s = 45 m. So, 0 - (12)^2 = 2 * a * 45, which gives -144 = 90a. Solving for a gives a = -1.6 m/s^2.
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