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Let's break down and solve the problem step by step. The question asks us to determine:

1. **Part A**: The velocity after 5.00 seconds.
2. **Part B**: The total distance traveled after 5.00 seconds.

The given data is:
- Initial velocity \( v_0 = 0 \) m/s (the object starts from rest),
- Acceleration \( a = 8.00 \, \text{m/s}^2 \),
- Time \( t = 5.00 \, \text{seconds} \).

### Part A: Finding the final velocity (\( v_f \)) after 5 seconds.

We can use the first equation of motion:

\[
v_f = v_0 + a \cdot t
\]

Substituting the known values:

\[
v_f = 0 + (8.00 \, \text{m/s}^2) \cdot (5.00 \, \text{seconds})
\]

\[
v_f = 40 \, \text{m/s}
\]

Thus, the final velocity after 5 seconds is \( v_f = 40 \, \text{m/s} \).

### Part B: Finding the total distance traveled after 5 seconds.

We can use the second equation of motion:

\[
x = v_0 \cdot t + \frac{1}{2} a \cdot t^2
\]

Substituting the known values:

\[
x = 0 + \frac{1}{2} \cdot (8.00 \, \text{m/s}^2) \cdot (5.00 \, \text{seconds})^2
\]

\[
x = \frac{1}{2} \cdot 8.00 \cdot 25 = 4.00 \cdot 25 = 100 \, \text{meters}
\]

Thus, the total distance traveled after 5 seconds is \( x = 100 \, \text{meters} \).

### Conclusion:
- The final velocity is \( 40 \, \text{m/s} \).
- The total distance traveled is \( 100 \, \text{meters} \).

Based on this, the correct answer is **option b**: \( v_f = 40 \, \text{m/s}, x = 100 \, \text{m} \).

---

Do you have any questions or need further details? Here are some related questions to explore:

1. What would the final velocity be after 10 seconds with the same acceleration?
2. How would the distance traveled change if the time were doubled?
3. What is the relationship between acceleration, time, and velocity in uniform acceleration?
4. How do we calculate distance when acceleration is not constant?
5. What happens to velocity and distance if the object starts with an initial velocity instead of from rest?

**Tip**: Always check the units when solving kinematic problems to ensure that they are consistent across all variables.

Un objeto parte del reposo con una aceleración constante de 8.00 m/s^2 a lo largo de una línea recta. Encuentre a) la rapidez después de 5.00 segundos y b) la distancia total recorrida en los 5.00 segundos.

Learn how to calculate the final velocity and total distance traveled by an object under constant acceleration of 8.00 m/s^2 over 5 seconds. Using the kinematic equations of motion, we solve for both velocity and distance. Suitable for high school students in Grades 10-12.

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