一个物体从静止开始做匀加速直线运动，加速度大小为2 m/s²。求物体在第3秒内的位移大小。
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Correct answer
B. 5 m
Principle or equation
匀变速直线运动位移公式：x = v0 t + (1/2) a t²。第3秒内的位移等于前3秒位移减去前2秒位移。
Why this answer is correct
前2秒位移 x2 = 0 + 0.5 × 2 × 2² = 4 m；前3秒位移 x3 = 0.5 × 2 × 3² = 9 m；第3秒内位移 Δx = x3 - x2 = 9 - 4 = 5 m。
Example
若加速度为3 m/s²，则前2秒位移为6 m，前3秒为13.5 m，第3秒内为7.5 m。
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