一质点的运动方程为x=4t+2,y=3t^2-6t+5,则质点速度最小的位置在?

求解题过程谢谢
2025-03-15 12:29:11
推荐回答(4个)
回答1:

x=4t+2

y=3t²-6t+5,

dx=4dt

dy=(6t-6)dt=6(t-1)dt

v=√{(dx)²+(dy)²}/dt 

= √{4²+6²(t-1)²} 

= 2√{4+9t²-18t+9}

= 2√{9(t-1)²+4}

t=1时速度最小值vmin=2√(0+4)=4

x=4t+2=6

y=3t²-6t+5=3-6+5=2

应为(6,2)

扩展资料

1、物理上的速度是一个相对量,即一个物体相对另一个物体(参照物)位移在单位时间内变化的的大小。

2、物理上还有平均速度:物体通过一段位移和所用时间的比值为物体在该位移的平均速度,平时我们说的多是瞬时速度。

3、平时我们形容单位时间做的某种动作的快慢或多少时也会用到速度。比如:打字速度、翻译速度。

4、速度是矢量,无论平均速度还是瞬时速度都是矢量。区分速度与速率的唯一标准就是速度有大小也有方向,速率则有大小没方向。

回答2:


答案如图所示

回答3:


以上解答

回答4:

下面我的解答思路和过程
质点的运动方程(轨迹)是:为x=4t+2; y=3t^2-6t+5。
则,质点在x , y 方向的运动速度分别为:Vx=dx/dt =4, Vy=dy/dt =6t-6.
则质点的总运动速度V(不标记运动方向的标量)是Vx 与 Vy 的合成,
即 V =√{(Vx)^2 + (Vy)^2} =√(36t^2 - 72t +52)
由 dV/dt =0 求V极限值时的t,则求有72t-72=0, t=1;
将此t=1带入质点运动轨迹方程,可得x=6, y=2.
即,质点速度最小的位置在(6, 2)

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