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We are provided with an acceleration-time graph and asked to find the total distance traveled by the car during the first 26 seconds of its journey.

### Step-by-step Analysis:

1. **Given Information:**
   - The acceleration graph consists of three distinct segments:
     - From \( t = 0 \) to \( t = 4 \) seconds, the acceleration is constant at \( 1.5 \, \text{m/s}^2 \).
     - From \( t = 4 \) to \( t = 24 \) seconds, the acceleration is zero (meaning constant velocity).
     - From \( t = 24 \) to \( t = 26 \) seconds, the acceleration is \( -4 \, \text{m/s}^2 \) (deceleration).

   - The distance traveled in the first 4 seconds is given as \( 24 \, \text{m} \).
   
2. **First Segment (0 to 4 seconds):**
   - Using the given information, the distance traveled in this period is \( 24 \, \text{m} \).

3. **Second Segment (4 to 24 seconds):**
   - The velocity at \( t = 4 \) seconds can be found using the formula:
     \[
     v = u + at
     \]
     where \( u = 0 \, \text{m/s} \), \( a = 1.5 \, \text{m/s}^2 \), and \( t = 4 \, \text{s} \).
     \[
     v_4 = 0 + 1.5 \times 4 = 6 \, \text{m/s}
     \]
   - During the next 20 seconds (from \( t = 4 \) to \( t = 24 \)), the car moves with constant velocity \( v_4 = 6 \, \text{m/s} \).
   - The distance traveled during this phase is:
     \[
     d_2 = v_4 \times 20 = 6 \times 20 = 120 \, \text{m}
     \]

4. **Third Segment (24 to 26 seconds):**
   - Here, the car decelerates with \( a = -4 \, \text{m/s}^2 \) for 2 seconds.
   - The velocity at \( t = 24 \) seconds is still \( 6 \, \text{m/s} \).
   - The final velocity at \( t = 26 \) seconds can be calculated as:
     \[
     v_{26} = v_{24} + at = 6 + (-4) \times 2 = 6 - 8 = -2 \, \text{m/s}
     \]
   - The distance traveled during this deceleration is given by:
     \[
     d_3 = v_{24} \times t + \frac{1}{2} a t^2
     \]
     Substituting \( v_{24} = 6 \, \text{m/s} \), \( a = -4 \, \text{m/s}^2 \), and \( t = 2 \, \text{s} \):
     \[
     d_3 = 6 \times 2 + \frac{1}{2} \times (-4) \times 2^2 = 12 - 8 = 4 \, \text{m}
     \]

5. **Total Distance:**
   - The total distance traveled is the sum of all three segments:
     \[
     \text{Total Distance} = d_1 + d_2 + d_3 = 24 + 120 + 4 = 148 \, \text{m}
     \]

Thus, the total distance traveled by the car during the first 26 seconds of its journey is **148 meters**.

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Would you like further details or have any questions?

Here are five related questions to explore:
1. How would the graph change if the car started with an initial velocity?
2. What would the total distance be if the car decelerated more sharply after 24 seconds?
3. Can you determine the total displacement rather than distance using this graph?
4. How does constant acceleration affect the velocity and distance covered by an object?
5. How would the result differ if the car was moving uphill or against resistance?

**Tip:** Always remember that the area under an acceleration-time graph gives the change in velocity over time!

The acceleration-time graph shown in Figure 5 represents part of a journey made by a car along a straight horizontal road. The car accelerated from rest at time t = 0. The distance travelled by the car during the first 4 seconds of its journey has been found to be 24 m. Find the total distance travelled by the car during the first 26 s of its journey.

This problem involves calculating the total distance traveled by a car over 26 seconds using its acceleration-time graph. Learn how to apply kinematics equations and interpret acceleration-time graphs to find distances in a step-by-step solution. Suitable for students in Grades 10-12.

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