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To solve this problem, we can use the **principle of conservation of energy**. Here's the step-by-step approach:

1. **Understanding the Initial Scenario**:  
   When the car is nudged down the hill from rest (initial speed = 0 m/s), it reaches a final speed of 4 m/s. Since the only force doing work here is gravity, we assume all potential energy lost is converted into kinetic energy.

2. **Calculating the Initial Energy Transformation**:  
   Given that the car is 2 kg and its final speed after descending the hill from rest is 4 m/s, we can calculate the change in gravitational potential energy (PE) required to reach that speed.  
   
   The kinetic energy (KE) at the bottom when speed is 4 m/s is:
   \[
   KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 2 \, \text{kg} \times (4 \, \text{m/s})^2 = 16 \, \text{J}
   \]
   Therefore, the change in potential energy, \(\Delta PE\), is also 16 J (since it all converts to kinetic energy).

3. **Applying This Information with an Initial Speed of 3 m/s**:  
   Now, if the car starts with an initial speed of 3 m/s, it already has some initial kinetic energy:
   \[
   KE_{\text{initial}} = \frac{1}{2}mv^2 = \frac{1}{2} \times 2 \, \text{kg} \times (3 \, \text{m/s})^2 = 9 \, \text{J}
   \]
   As it moves down the hill, it will gain an additional 16 J from the potential energy change.

4. **Calculating the Final Kinetic Energy and Speed**:  
   The total kinetic energy at the bottom will now be:
   \[
   KE_{\text{total}} = KE_{\text{initial}} + \Delta PE = 9 \, \text{J} + 16 \, \text{J} = 25 \, \text{J}
   \]
   
   To find the final speed, \(v_f\), we use the kinetic energy formula:
   \[
   KE_{\text{total}} = \frac{1}{2}mv_f^2
   \]
   Solving for \(v_f\):
   \[
   v_f = \sqrt{\frac{2 \cdot KE_{\text{total}}}{m}} = \sqrt{\frac{2 \cdot 25}{2}} = \sqrt{25} = 5 \, \text{m/s}
   \]

So, the final speed of the car when it starts with an initial speed of 3 m/s will be **5 m/s**.

Would you like further details on the energy conservation method or other related concepts?

Here are five related questions to explore:

1. How would the answer change if frictional forces were present?
2. How is gravitational potential energy calculated in this scenario?
3. What is the significance of energy conservation in physics?
4. How does initial kinetic energy affect the final speed in such scenarios?
5. What other methods could we use to approach similar energy problems?

**Tip:** When dealing with energy problems on slopes or hills, always check if potential energy changes are converted fully into kinetic energy, which simplifies calculations under ideal conditions.

When nudged down the hill from stationary, the final speed of a 2kg car is 4 ms⁻¹. With an initial speed of 3 ms⁻¹, what is the final speed?

Learn how to calculate the final speed of a 2kg car on a hill using energy conservation principles. This problem involves potential and kinetic energy concepts, ideal for students in Grades 10-12. Explore how initial speed affects the final outcome.

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