一辆汽车从静止开始，在水平路面上以4m/s2的加速度做匀加速直线运动．求：
（1）2s末汽车速度的大小；
（2）2s末汽车位移的大小．
（1）根据匀变速直线运动的速度公式v=v0+at
则2s末汽车的速度大小为v=at=4×2m/s=8m/s．
（2）根据匀变速直线运动的位移公式x=v0t+
at21 2
则2s末汽车的位移的大小为x=
at2=1 2
×4×22m=8m．1 2
答：（1）2s末汽车速度的大小为8m/s；
（2）2s末汽车位移的大小为8m．
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