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
Correct answer
Explanation
The correct answer is C, −1.6 m/s². Using v² = u² + 2as with u = 12 m/s, v = 0 and s = 45 m gives a = −144 ÷ 90 = −1.6 m/s².
The third equation of motion links velocity and distance without needing time: v² = u² + 2as. Here the final velocity is zero because the car stops, so 0 = 144 + 90a, and a = −1.6 metres per second squared. The minus sign shows that the acceleration acts opposite to the motion, which is retardation or deceleration; in words the car slows down by 1.6 m/s every second. As a check, the time taken is 12 ÷ 1.6 = 7.5 s, and the average speed of 6 m/s over 7.5 s gives exactly 45 m.
A is wrong because a positive value would mean the car is speeding up, not stopping. B is wrong because 3.2 comes from forgetting the factor 2, and its sign is also wrong. D is wrong because 0.8 comes from dividing by 180 instead of 90.
Exam tip: Stopping problems use v² = u² + 2as with v = 0; the answer is always negative.
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