https://file.aiquickdraw.com/mathbot/file/1726388441709lwnkcbjl.jpg

The problem in the image states:

- A car has an initial velocity \( u = 2 \, \text{ms}^{-1} \).
- The car's acceleration is given by \( a(t) = \frac{5}{2} - \frac{t}{8} \, \text{ms}^{-2} \), where \( t \) is time.
- We are asked to calculate the distance traveled during the first four seconds.

To solve this problem, we can use kinematic equations, but since the acceleration is time-dependent, we'll need to integrate it to find velocity and then integrate again to find the distance.

### Step-by-Step Solution:

1. **Find the velocity**:
   - Velocity is the integral of acceleration with respect to time:
     \[
     v(t) = \int a(t) \, dt = \int \left( \frac{5}{2} - \frac{t}{8} \right) dt
     \]
   - Performing the integration:
     \[
     v(t) = \frac{5}{2}t - \frac{t^2}{16} + C
     \]
   - To find the constant \( C \), we use the initial condition that the initial velocity is \( u = 2 \, \text{ms}^{-1} \) at \( t = 0 \):
     \[
     v(0) = 2 \quad \Rightarrow \quad C = 2
     \]
     So the velocity function is:
     \[
     v(t) = \frac{5}{2}t - \frac{t^2}{16} + 2
     \]

2. **Find the distance**:
   - Distance (displacement) is the integral of velocity with respect to time:
     \[
     s(t) = \int v(t) \, dt = \int \left( \frac{5}{2}t - \frac{t^2}{16} + 2 \right) dt
     \]
   - Performing the integration:
     \[
     s(t) = \frac{5}{4}t^2 - \frac{t^3}{48} + 2t + D
     \]
   - Since the car starts from the origin, \( s(0) = 0 \), so the constant \( D = 0 \).
     Thus, the distance function is:
     \[
     s(t) = \frac{5}{4}t^2 - \frac{t^3}{48} + 2t
     \]

3. **Calculate the distance at \( t = 4 \) seconds**:
   - Substitute \( t = 4 \) into the equation for distance:
     \[
     s(4) = \frac{5}{4}(4)^2 - \frac{(4)^3}{48} + 2(4)
     \]
     \[
     s(4) = \frac{5}{4} \times 16 - \frac{64}{48} + 8
     \]
     \[
     s(4) = 20 - \frac{4}{3} + 8
     \]
     \[
     s(4) = 28 - \frac{4}{3} = \frac{84}{3} - \frac{4}{3} = \frac{80}{3}
     \]
     \[
     s(4) \approx 26.67 \, \text{meters}
     \]

So, the car travels approximately **26.67 meters** in the first 4 seconds.

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Would you like more details on any step or have any questions?

Here are 5 related questions you might find interesting:
1. How would the distance change if the initial velocity was different?
2. How do you calculate the maximum velocity of the car?
3. How does the time-dependent acceleration affect the motion compared to constant acceleration?
4. What would the displacement be if the time was extended to 6 seconds?
5. How would you calculate the average velocity over the 4 seconds?

**Tip**: In kinematics, integrating the acceleration function gives you velocity, and integrating velocity gives you displacement.

A car with initial velocity 2 m/s moves in a straight line with acceleration a = (5/2 - t/8) m/s^2. Calculate the distance traveled during the first four seconds.

This problem involves calculating the distance traveled by a car with initial velocity 2 m/s and time-dependent acceleration a = (5/2 - t/8) m/s² during the first four seconds. The solution uses kinematic equations, integration of the acceleration, and velocity functions.

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