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Left closed in review as "Original close reason(s) were not resolved" by Amit, Miyase, John Rennie
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If the velocityspeed distribution, in m/s, of a flow is given by $( v = 2x^3 + 2y^2 - 3z )$$v = 2x^3 + 2y^2 - 3z$, then the acceleration of the fluid at the point with coordinates $([2, 1, 5])$$[2, 1, 5]$, in meters, will be greater than $20, \text{m/s}^2 $.

How can I prove that?

I tried to apply the definition of material derivative by expressing velocity in terms of magnitude and a unit vector, but I was unsuccessful. I suspect that some form of maximization must be applied, but I don't know which one. Because, depending on the direction of velocity, acceleration changes but remains restricted to a range of values.

If the velocity distribution, in m/s, of a flow is given by $( v = 2x^3 + 2y^2 - 3z )$, then the acceleration of the fluid at the point with coordinates $([2, 1, 5])$, in meters, will be greater than $20, \text{m/s}^2 $.

How can I prove that?

I tried to apply the definition of material derivative by expressing velocity in terms of magnitude and a unit vector, but I was unsuccessful. I suspect that some form of maximization must be applied, but I don't know which one. Because, depending on the direction of velocity, acceleration changes but remains restricted to a range of values.

If the speed distribution, in m/s, of a flow is given by $v = 2x^3 + 2y^2 - 3z$, then the acceleration of the fluid at the point with coordinates $[2, 1, 5]$, in meters, will be greater than $20, \text{m/s}^2 $.

How can I prove that?

I tried to apply the definition of material derivative by expressing velocity in terms of magnitude and a unit vector, but I was unsuccessful. I suspect that some form of maximization must be applied, but I don't know which one. Because, depending on the direction of velocity, acceleration changes but remains restricted to a range of values.

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Amit
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If the velocity distribution, in m/s, of a flow is given by $( v = 2x^3 + 2y^2 - 3z )$, then the acceleration of the fluid at the point with coordinates $([2, 1, 5])$, in meters, will be greater than $20, \text{m/s}^2 $.

How can I proofprove that?

I tried to apply the definition of material derivative by expressing velocity in terms of magnitude and a unit vector, but I was unsuccessful. I suspect that some form of maximization must be applied, but I don't know which one. Because, depending on the direction of velocity, acceleration changes but remains restricted to a range of values.

If the velocity distribution, in m/s, of a flow is given by $( v = 2x^3 + 2y^2 - 3z )$, then the acceleration of the fluid at the point with coordinates $([2, 1, 5])$, in meters, will be greater than $20, \text{m/s}^2 $.

How can I proof that?

I tried to apply the definition of material derivative by expressing velocity in terms of magnitude and a unit vector, but I was unsuccessful. I suspect that some form of maximization must be applied, but I don't know which one. Because, depending on the direction of velocity, acceleration changes but remains restricted to a range of values.

If the velocity distribution, in m/s, of a flow is given by $( v = 2x^3 + 2y^2 - 3z )$, then the acceleration of the fluid at the point with coordinates $([2, 1, 5])$, in meters, will be greater than $20, \text{m/s}^2 $.

How can I prove that?

I tried to apply the definition of material derivative by expressing velocity in terms of magnitude and a unit vector, but I was unsuccessful. I suspect that some form of maximization must be applied, but I don't know which one. Because, depending on the direction of velocity, acceleration changes but remains restricted to a range of values.

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