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Chapter 6: Problem 2

The velocity in a certain flow field is given by the equation $$\mathbf{v}=x \hat{\mathbf{i}}+x^{2} z \hat{\mathbf{j}}+y z \hat{\mathbf{k}}$$ Determine the expressions for the three rectangular components of acceleration.

Short Answer

Expert verified
The components of the acceleration vector are: \(a_x = 0\), \(a_y = x^{2} \cdot \frac{{dz}}{{dt}} + z \cdot \frac{{d(x^{2})}}{{dt}}\), \(a_z = y \cdot \frac{{dz}}{{dt}} + z \cdot \frac{{dy}}{{dt}} \).

Step by step solution

01

Derive the x-component of acceleration

The x-component of velocity is \(x\). To obtain the x-component of acceleration (\(a_x\)), compute the derivative of velocity with respect to time namely: \(a_x = \frac{{dv_x}}{{dt}}\). But there is no t in the expression, so the derivative of x becomes zero as it's not a function of t.
02

Derive the y-component of acceleration

The y-component of velocity is given by \(x^{2} \cdot z\). Obtain \(a_y\) by taking the derivative of it. Since it's composed of two functions x and z, the product rule is applied, i.e., \(a_y = \frac{{dv_y}}{{dt}} = x^{2} \cdot \frac{{dz}}{{dt}} + z \cdot \frac{{d(x^{2})}}{{dt}}\). Here, \(\frac{{dz}}{{dt}}\) and \(\frac{{d(x^{2})}}{{dt}}\) need to be replaced with their corresponding velocity components.
03

Derive the z-component of acceleration

The z-component of velocity is given by \(y \cdot z\). Obtain \(a_z\) by taking the derivative of it. Similar to the y component, since it's composed of two functions y and z, the product rule is applied, i.e., \(a_z = \frac{{dv_z}}{{dt}} = y \cdot \frac{{dz}}{{dt}} + z \cdot \frac{{dy}}{{dt}} \). Here, \(\frac{{dz}}{{dt}}\) and \(\frac{{dy}}{{dt}}\) need to be replaced with their corresponding velocity components.

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Most popular questions from this chapter

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