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The image provided contains a physics problem related to motion and distance. Here’s a breakdown of the problem:

**Problem Statement:**

- A car moves from rest along a straight line for 20 seconds. It accelerates at a constant rate.

- **Given Information:**
  - The total distance covered by the car, \( s(t) \), is given by the equation:
    \[
    s(t) = 0.8 \times t \times \alpha \times t^2
    \]
    where \( \alpha \) is a constant and \( t \) is the time in seconds.

- **Question:**
  - What is the value of the distance traveled?

**Solution Outline:**

To solve this, follow these steps:

1. **Understand the equation:** The equation given is:
   \[
   s(t) = 0.8 \times t \times \alpha \times t^2
   \]
   which can be rewritten as:
   \[
   s(t) = 0.8 \alpha t^3
   \]
   
2. **Determine the value of \( \alpha \) and other constants:** 
   - Since the problem does not explicitly state the value of \( \alpha \), typically, \( \alpha \) might represent the acceleration or another constant. However, without this value, it is assumed that the problem might require symbolic or dimensional analysis, or \( \alpha \) might be provided in an earlier part of a larger problem set.

3. **Calculate the distance traveled:** 
   - If \( \alpha \) is known, you substitute \( t = 20 \) seconds into the equation:
     \[
     s(20) = 0.8 \times \alpha \times (20)^3 = 0.8 \times \alpha \times 8000 = 6400 \alpha
     \]
   - Therefore, the distance traveled in 20 seconds is \( 6400 \alpha \) meters.

Without knowing \( \alpha \), the answer is left in terms of \( \alpha \).

If you have any additional information about \( \alpha \) or other aspects, feel free to share it. Would you like a more detailed exploration or explanation of any step?

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**Next Steps and Related Questions:**

1. What physical meaning might the constant \( \alpha \) have in this context?
2. How would the distance change if the acceleration were doubled?
3. What assumptions are made in using this equation for motion?
4. How would you determine \( \alpha \) if given initial and final velocities?
5. How does the equation relate to the standard kinematic equation \( s = ut + \frac{1}{2}at^2 \)?

**Tip:** When solving problems involving motion, always keep track of units, as they can guide you to correct interpretations and solutions.

A car moves from rest along a straight line for 20 seconds. It accelerates at a constant rate. The total distance covered by the car, s(t), is given by the equation: s(t) = 0.8 × t × α × t^2, where α is a constant and t is the time in seconds. What is the value of the distance traveled?

Explore how to calculate the distance traveled by a car accelerating at a constant rate using the equation s(t) = 0.8 × α × t^3. This problem covers key kinematic concepts and algebraic manipulation. Suitable for high school students in Grades 9-12.

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