To solve this problem, we need to find the components of the net force on the airplane from the given momentum equation.

Given:

Momentum $\vec{p}(t)$ is provided as: $\vec{p}(t) = \left( -0.75 \, \text{kg} \cdot \text{m/s}^3 \right)t^2 \hat{i} + \left( 3.0 \, \text{kg} \cdot \text{m/s} \right) \hat{i} + \left( 0.25 \, \text{kg} \cdot \text{m/s}^2 \right)t \hat{j}$

Unknown:

The net force $\vec{F}(t)$, which has x-, y-, and z-components (if applicable).

Equation:

Force is related to momentum by the equation: $\vec{F}(t) = \frac{d\vec{p}(t)}{dt}$ Thus, we can find the net force by differentiating the momentum equation with respect to time.

Solution:

Let's differentiate each term of the momentum function:

1. For the $\hat{i}$ (x-component) term: = -0.75 \cdot 2t \, \text{kg} \cdot \text{m/s}^3 = -1.5t \, \text{N}$$ The constant term $$3.0 \, \text{kg} \cdot \text{m/s}$$ disappears since the derivative of a constant is zero.
2. For the $\hat{j}$ (y-component) term: $\frac{d}{dt}\left( 0.25 \, \text{kg} \cdot \text{m/s}^2 \cdot t \right) = 0.25 \, \text{kg} \cdot \text{m/s}^2 = 0.25 \, \text{N}$

Final Answer:

The net force $\vec{F}(t)$ is: $\vec{F}(t) = (-1.5t) \hat{i} + (0.25) \hat{j}$

Thus, the x-component of the net force is $-1.5t \, \text{N}$, the y-component is $0.25 \, \text{N}$, and there is no z-component (since the momentum expression doesn't include any $\hat{k}$-terms).

Do you have any questions, or would you like more details on the differentiation process?

Related Questions:

1. How does force relate to the rate of change of momentum in general?
2. What would happen if the momentum had a $z$-component in this problem?
3. How can you determine the acceleration of the airplane using the force found?
4. What physical meaning does the constant $3.0 \, \text{kg} \cdot \text{m/s}$ term have in the momentum equation?
5. How would you interpret the units of the various terms in the momentum equation?

Tip: Always make sure the units of momentum are consistent (e.g., kg·m/s) before performing any operations on them.