using the GRESA method, try answering the following: a car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 meters. Determine the acceleration of the car.

To solve the problem using the GRESA method, we break it down as follows:

### **GRESA Method:**

1. **Given:**
   - Initial velocity (\(v_i\)) = 0 m/s (since the car starts from rest)
   - Time (\(t\)) = 5.21 s
   - Distance (\(d\)) = 110 m

2. **Required:**
   - The acceleration (\(a\)) of the car

3. **Equation:**
   - We can use the second equation of motion that relates distance, initial velocity, time, and acceleration:
   \[
   d = v_i \cdot t + \frac{1}{2} a t^2
   \]
   Since \(v_i = 0\) (car starts from rest), the equation simplifies to:
   \[
   d = \frac{1}{2} a t^2
   \]

4. **Solution:**
   - Substitute the known values into the equation:
   \[
   110 = \frac{1}{2} a (5.21)^2
   \]
   - Solve for \(a\):
   \[
   110 = \frac{1}{2} a \cdot 27.1441
   \]
   \[
   110 = 13.57205a
   \]
   \[
   a = \frac{110}{13.57205} \approx 8.11 \text{ m/s}^2
   \]

5. **Answer:**
   - The acceleration of the car is approximately **8.11 m/s²**.

Would you like more details, or do you have any questions?

### **Relative Questions:**
1. What would happen to the acceleration if the distance were doubled?
2. How would the problem change if the car had an initial velocity of 5 m/s?
3. What formula should be used if the final velocity of the car is also given?
4. Can you derive the time if the acceleration and distance are known?
5. How would friction affect the acceleration of the car in a real-world scenario?

**Tip:** When dealing with uniformly accelerated motion, remember that the initial velocity plays a crucial role in determining the correct equation to use.

Learn how to calculate the acceleration of a car using the GRESA method with a detailed step-by-step solution. This problem involves uniformly accelerated motion, equations of motion, and practical application of physics principles. Suitable for students in Grades 10-12.

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