To solve for the net force exerted on the car, we need to use Newton's second law of motion:

$F_{\text{net}} = m \cdot a$

where:

• $F_{\text{net}}$ is the net force,
• $m$ is the mass of the car (730 kg),
• $a$ is the acceleration of the car.

First, we need to calculate the acceleration $a$ of the car. Since the car starts from rest and travels a distance of 39.8 meters in 3.01 seconds with uniform acceleration, we can use the kinematic equation:

$d = v_0 t + \frac{1}{2} a t^2$

where:

• $d = 39.8 \, \text{m}$ is the distance,
• $v_0 = 0 \, \text{m/s}$ (since the car starts from rest),
• $t = 3.01 \, \text{s}$,
• $a$ is the acceleration.

Simplifying the equation, since $v_0 = 0$:

$39.8 = \frac{1}{2} a (3.01)^2$

Now solve for $a$:

$39.8 = \frac{1}{2} a (9.0601)$ $39.8 = 4.53005 a$ $a = \frac{39.8}{4.53005} \approx 8.785 \, \text{m/s}^2$

Now that we have the acceleration, we can calculate the net force:

$F_{\text{net}} = 730 \cdot 8.785 \approx 6403.05 \, \text{N}$

So, the net force exerted on the car is approximately $6403 \, \text{N}$.

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Do you want more details on any step? Here are some related questions for further exploration:

1. How would the net force change if the car's mass increased by 10%?
2. What would be the final velocity of the car after 3.01 seconds?
3. If the car traveled 50 meters instead of 39.8 meters, how would the acceleration change?
4. What if the car had an initial velocity of 5 m/s? How would the equation change?
5. How can we find the work done by the net force over this distance?

Tip: When solving for acceleration with uniformly accelerated motion, the kinematic equations are a valuable tool to relate time, distance, and velocity.