一辆汽车从静止开始做匀加速直线运动，加速度大小为2 m/s²。汽车在最初2秒内的位移大小是多少？
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Correct answer
B. 4 m
Principle or equation
匀变速直线运动位移公式：s = v₀t + ½at²。初速度v₀=0，a=2 m/s²，t=2 s。
Why this answer is correct
代入公式：s = 0×2 + ½×2×2² = 0 + ½×2×4 = 4 m。因此位移为4 m。
Example
若加速度为3 m/s²，时间3 s，则位移s=½×3×9=13.5 m。
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