初速10m/s,关油门后前进150m,速度减为5m/s,求再过30s,前进的距离?若用 150=10+1/2at^2 ,5=10-at 怎么算,和正确答案不一样啊就是用x=v+1/2at^2 vt=v0+at
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![初速10m/s,关油门后前进150m,速度减为5m/s,求再过30s,前进的距离?若用 150=10+1/2at^2 ,5=10-at 怎么算,和正确答案不一样啊就是用x=v+1/2at^2 vt=v0+at](/uploads/image/z/3377553-33-3.jpg?t=%E5%88%9D%E9%80%9F10m%EF%BC%8Fs%2C%E5%85%B3%E6%B2%B9%E9%97%A8%E5%90%8E%E5%89%8D%E8%BF%9B150m%2C%E9%80%9F%E5%BA%A6%E5%87%8F%E4%B8%BA5m%2Fs%2C%E6%B1%82%E5%86%8D%E8%BF%8730s%2C%E5%89%8D%E8%BF%9B%E7%9A%84%E8%B7%9D%E7%A6%BB%3F%E8%8B%A5%E7%94%A8%E3%80%80%E3%80%80150%3D10%2B1%2F2at%5E2+%E3%80%80%E3%80%80%2C5%3D10-at+%E6%80%8E%E4%B9%88%E7%AE%97%2C%E5%92%8C%E6%AD%A3%E7%A1%AE%E7%AD%94%E6%A1%88%E4%B8%8D%E4%B8%80%E6%A0%B7%E5%95%8A%E5%B0%B1%E6%98%AF%E7%94%A8x%3Dv%2B1%2F2at%5E2+vt%3Dv0%2Bat)
初速10m/s,关油门后前进150m,速度减为5m/s,求再过30s,前进的距离?若用 150=10+1/2at^2 ,5=10-at 怎么算,和正确答案不一样啊就是用x=v+1/2at^2 vt=v0+at
初速10m/s,关油门后前进150m,速度减为5m/s,求再过30s,前进的距离?
若用 150=10+1/2at^2 ,5=10-at 怎么算,和正确答案不一样啊
就是用x=v+1/2at^2 vt=v0+at
初速10m/s,关油门后前进150m,速度减为5m/s,求再过30s,前进的距离?若用 150=10+1/2at^2 ,5=10-at 怎么算,和正确答案不一样啊就是用x=v+1/2at^2 vt=v0+at
V'² - V²=2aS
5² - 10²=2aX150
加速度: a=-0.25m/s²
vt=v+at
0=5-0.25t
t=20s (20s运动, 10S停止了)
再过30s,前进的距离:S=20X(5+0)/2=50m