额定功率启动过程--速度与位移确切关系

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1、汽车以额定功率启动过程－－速度和位移确切关系由牛顿第二定律：，整理得：；由初始条件：当t=0时，V＝0代入得即此为速度与时间的关系。由动能定理：；把上面的t代入，可得sP分析时间：;;可得，再用表示达到最大速度的比例，则；其中t0为由Vm减速至停止所用时间再分析位移：＝令;则。令；；则,.0.0400.000820.000020.7200.552970.293770.9802.932022.451820.0800.003380.000180.7400.607070.333270.9812.982322.501140.1200.007830.0006

2、30.7600.667120.378320.9823.035392.553220.1600.014350.001550.7800.734130.429930.9833.091542.608400.2000.023140.003140.8000.809440.489440.9843.151172.667040.2400.034440.005640.8100.850730.522680.9853.214712.729590.2800.048500.009300.8200.894800.558600.9863.282702.796600.3200.065

3、660.014460.8300.941960.597510.9873.355812.868720.3600.086290.021490.8400.992580.639780.9883.434852.946780.4000.110830.030830.8501.047120.685870.9893.520863.031800.4400.139820.043020.8601.106110.736310.9903.615173.125120.4800.173930.058730.8701.170220.791770.9913.719533.228490.

4、5000.193150.061850.8801.240260.853060.9923.836323.344280.5200.213970.078770.8901.317280.921230.9933.968853.475820.5400.236530.090730.9001.402590.997590.9944.122003.627980.5600.260980.104180.9101.497951.083900.9954.303323.808310.5800.287500.119300.9201.605731.182530.9964.525474

5、.029460.6000.316290.136290.9301.729261.296810.9974.812154.315150.6200.347580.155380.9401.873411.431610.9985.216624.718620.6400.381650.176850.9502.045731.594480.9995.908785.409780.6600.418810.201010.9602.258881.798080.6800.459430.228230.9702.536562.066110.7000.503970.258970.980

6、2.932022.45182例：汽车发动机的额定功率为80KW，汽车的质量为2000kg，汽车在水平路面上行驶时，阻力是车重的0.1倍，g=10m/s2。汽车保持额定功率不变从静止启动后:(1)汽车所能达到的最大速度是多大？(2)当汽车的加速度为3m/s2时速度多大？这两个问题高中学生就可求出！(3)当汽车的速度为20m/s时，经历了多少时间？走了的位移为多少？(4)当汽车的走的位移为2000m时，用了多少时间？速度达到多少？解：（1）当时间很长后，汽车可看作以最大速度匀速运动、此时F＝fP=fVm,即Vm=P/f=40m/s(2)P=Fv;F-f

7、=ma可得＝10m/s(3)，查上面的表得：，,,(4),查上面的表得，0.927；1.69，得t=t0=67.6s;V＝Vm=37.1s