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A 5T car, the rated power of the engine is 80kW, the car from static start to acceleration of 1 meter per second to do uniform acceleration linear motion, the car resistance It is 0.06 times of the vehicle weight, G is taken as 10, and the maximum time for the vehicle to make a uniform acceleration straight line? The instantaneous power at the end of 5S and 15 seconds after the vehicle starts to move?
If f-f = ma, then f = F + Ma = 0.06mg + Ma = 8000n. According to the power P = FV, when p = PE (PE is the rated power), the power reaches the maximum value, and the vehicle speed will reach the maximum speed in the uniform acceleration stage. We set V1, V1 = 10m / s. therefore, from vt = V0 + at, t = V1 / a = 10s
At the end of 5S, the vehicle is in the stage of uniform acceleration, and the power has not reached the rated power. At this time, the instantaneous power can be calculated by P = FV. The speed at the end of 5S is set as V2, V2 = at2 = 1 * 5 = 5m / s. the instantaneous power P = fv2 = 8000 * 5 = 40000 W
At the end of 10s, the power of the car has reached the maximum rated power, and then it will keep the rated power and drive all the time, so the instantaneous power at the end of 15s is p = PE = 80000 W
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