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。求物体在前3秒内的位移大小。
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Correct answer
B. 9 m
Principle or equation
匀变速直线运动位移公式：x = v₀t + ½at²。初速度v₀=0，a=2 m/s²，t=3 s。
Why this answer is correct
代入公式：x = 0 + ½ × 2 × 3² = ½ × 2 × 9 = 9 m。
Example
若加速度为3 m/s²，时间2 s，则位移 = ½ × 3 × 2² = 6 m。
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