一物体从静止开始做匀加速直线运动，加速度大小为2 m/s²。求该物体在第3秒内的位移大小。
Show the answer and explanation
Correct answer
B. 5 m
Principle or equation
第n秒内的位移等于前n秒位移减去前(n-1)秒位移。匀加速直线运动位移公式：s = (1/2) a t²。
Why this answer is correct
前3秒位移 s3 = (1/2)*2*3² = 9 m；前2秒位移 s2 = (1/2)*2*2² = 4 m；第3秒内位移 Δs = s3 - s2 = 9 - 4 = 5 m。
Example
若加速度为3 m/s²，求第2秒内位移：s2 = 1.5*4=6 m，s1=1.5*1=1.5 m，Δs=4.5 m。
This published item includes a stored explanation and passed the platform’s publication workflow. It is independent preparation material, not a claim of an official or recalled examination question.