If a car converts 1.2 X 106 J of chemical energy into motion by speeding up from 25 mph to 50 mph, how much chemical energy would be needed to speed up from 50 mph to 75 mph, assuming the same efficiency?

To find the chemical energy needed for the car to speed up from 50 mph to 75 mph, we’ll need to understand the relationship between kinetic energy and speed. 

### Steps to Solve:
1. **Calculate Initial Kinetic Energy:**  
   The kinetic energy (\( KE \)) of an object is given by the formula:  
   \[
   KE = \frac{1}{2}mv^2
   \]
   where \( m \) is the mass and \( v \) is the velocity.

2. **Set Up the Relationship:**  
   The car’s speed is increasing in two stages:
   - From \( 25 \, \text{mph} \) to \( 50 \, \text{mph} \).
   - From \( 50 \, \text{mph} \) to \( 75 \, \text{mph} \).

3. **Kinetic Energy for Each Stage:**  
   Let’s denote the car's mass as \( m \).  
   
   - For the first stage (from \( v_1 = 25 \, \text{mph} \) to \( v_2 = 50 \, \text{mph} \)), the change in kinetic energy is given by:  
     \[
     \Delta KE_1 = \frac{1}{2}m(v_2^2 - v_1^2)
     \]
     Since this requires \( 1.2 \times 10^6 \, \text{J} \), we can write:
     \[
     1.2 \times 10^6 = \frac{1}{2}m(50^2 - 25^2)
     \]
   
   - For the second stage (from \( v_3 = 50 \, \text{mph} \) to \( v_4 = 75 \, \text{mph} \)), the change in kinetic energy would be:  
     \[
     \Delta KE_2 = \frac{1}{2}m(v_4^2 - v_3^2)
     \]
     \[
     = \frac{1}{2}m(75^2 - 50^2)
     \]

4. **Relating the Two Energy Changes:**
   Since \( m \) cancels out in a proportion, we can determine the ratio of the two kinetic energy changes:
   \[
   \frac{\Delta KE_2}{\Delta KE_1} = \frac{75^2 - 50^2}{50^2 - 25^2}
   \]
   
   **Calculations:**  
   \[
   75^2 - 50^2 = 5625 - 2500 = 3125
   \]
   \[
   50^2 - 25^2 = 2500 - 625 = 1875
   \]
   Therefore,
   \[
   \frac{\Delta KE_2}{\Delta KE_1} = \frac{3125}{1875} = \frac{5}{3}
   \]
   
5. **Solving for \(\Delta KE_2\):**  
   Now we can find the chemical energy needed for the second stage:
   \[
   \Delta KE_2 = \frac{5}{3} \times (1.2 \times 10^6) = 2 \times 10^6 \, \text{J}
   \]

### Conclusion
The car would require \( 2 \times 10^6 \, \text{J} \) of chemical energy to speed up from 50 mph to 75 mph.

Would you like a more detailed breakdown, or do you have further questions?

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#### Additional Related Questions:
1. What would be the chemical energy needed to accelerate from 75 mph to 100 mph?
2. How does the required energy change if the mass of the car doubles?
3. If the car's engine has 75% efficiency, what total chemical energy would be required?
4. How would the required energy change if we included air resistance?
5. What formula would you use if the car was moving up a hill at a constant incline?

#### Tip:
Remember, the energy required for acceleration increases quadratically with speed, so doubling the speed quadruples the required kinetic energy!

If a car converts 1.2 × 10^6 J of chemical energy into motion by speeding up from 25 mph to 50 mph, how much chemical energy would be needed to speed up from 50 mph to 75 mph, assuming the same efficiency?

Find out how much chemical energy is needed to speed up a car from 50 mph to 75 mph. This problem covers concepts in kinetic energy and proportional relationships, using detailed formulas to calculate energy changes. Suitable for high school students in Grades 9-12.

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