When a particle moves along the x-axis, the relationship between its acceleration a and the position coordinate X is a = 2 + 6x2 (a is equal to the square of 2 plus 6 x), if the velocity of the particle at the origin is zero Try to find the speed at any position. (I can't calculate the second-order differential, please write in detail, I'm stupid.)
1.dv/dt=2+6x2
2.dx/dt=v
The second formula is written as DT = DX / V and substituted into the form to get: VDV = (2 + 6x2) DX
Then integral, too lazy to calculate
If you can't, read the book
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