The average power generated by the force is approximately \(35,742.7 \, \text{W}\) or \(35.7 \, \text{kW}\).
The average power (\(P_{\text{avg}}\)) can be calculated using the formula:
\[ P_{\text{avg}} = \frac{W}{\Delta t} \]
where \(W\) is the work done and \(\Delta t\) is the time interval.
The work done (\(W\)) can be expressed as the product of force (\(F\)), displacement (\(d\)), and the cosine of the angle (\(\theta\)) between the force and displacement vectors:
\[ W = F \cdot d \cdot \cos(\theta) \]
In this case, the force is causing the car to accelerate in the x-direction, and the displacement is given by the equation of motion:
\[ d = \frac{1}{2} a t^2 \]
where \(a\) is the acceleration and \(t\) is the time.
Now, let's calculate the displacement and work done:
\[ d = \frac{1}{2} \cdot 4.20 \, \text{m/s}^2 \cdot (3.70 \, \text{s})^2 \]
\[ d = \frac{1}{2} \cdot 4.20 \, \text{m/s}^2 \cdot 13.69 \, \text{s}^2 \]
\[ d = 28.665 \, \text{m} \]
Now, calculate the work done:
\[ W = F \cdot d \cdot \cos(\theta) \]
Since the force is causing the motion in the same direction as the displacement, \(\cos(\theta) = 1\), and the equation simplifies to:
\[ W = F \cdot d \]
The force can be determined using Newton's second law (\(F = ma\)), where \(m\) is the mass and \(a\) is the acceleration:
\[ F = m \cdot a \]
\[ F = (1.10 \times 10^3 \, \text{kg}) \cdot (4.20 \, \text{m/s}^2) \]
\[ F = 4620 \, \text{N} \]
Now, calculate the work done:
\[ W = F \cdot d \]
\[ W = (4620 \, \text{N}) \cdot (28.665 \, \text{m}) \]
\[ W = 132376.2 \, \text{J} \]
Finally, calculate the average power:
\[ P_{\text{avg}} = \frac{W}{\Delta t} \]
\[ P_{\text{avg}} = \frac{132376.2 \, \text{J}}{3.70 \, \text{s}} \]
\[ P_{\text{avg}} \approx 35,742.7 \, \text{W} \]
So, the average power generated by the force is approximately \(35,742.7 \, \text{W}\) or \(35.7 \, \text{kW}\).