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/s²
- B.+3.2 m/s²
- C.−1.6 m/s²
- D.−0.8 m/s²
Show answer
Using v² = u² + 2as with u = 12 m/s, v = 0 and s = 45 m gives a = −144 ÷ 90 = −1.6 m/s².
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