To solve the problem of how much power is required to accelerate the car, we can follow these steps:

Given:

• Mass of the car, $m = 1430 \, \text{kg}$
• Initial velocity, $v_1 = 7.50 \, \text{m/s}$
• Final velocity, $v_2 = 11.0 \, \text{m/s}$
• Time, $t = 9.30 \, \text{s}$

Step 1: Calculate the change in kinetic energy ($\Delta KE$).

Kinetic energy is given by the formula: $KE = \frac{1}{2} m v^2$ The change in kinetic energy is: $\Delta KE = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2$ Substituting the given values: $\Delta KE = \frac{1}{2} \times 1430 \times (11.0^2 - 7.50^2) \, \text{J}$ $\Delta KE = 715 \times (121 - 56.25) \, \text{J}$ $\Delta KE = 715 \times 64.75 \, \text{J}$ $\Delta KE = 46391.25 \, \text{J}$

Step 2: Calculate the power.

Power is the rate at which work is done, which is the change in energy over time: $P = \frac{\Delta KE}{t}$ Substituting the values: $P = \frac{46391.25}{9.30} \, \text{W}$ $P \approx 4987.23 \, \text{W}$

Thus, the power required is approximately 4987 W or 4.99 kW.

Would you like a more detailed breakdown or have any questions?

Here are five related questions to expand on this topic:

1. How is kinetic energy related to the work-energy theorem?
2. What factors affect the amount of power needed to accelerate a vehicle?
3. How would including friction change the calculation of power?
4. What are the units of power and how are they related to energy and time?
5. Can power be negative, and under what circumstances?

Tip: Always remember that power is a measure of the rate at which work is done or energy is transferred, and it is directly proportional to both the force applied and the velocity at which the force is applied.