We are given the position function of an object moving along a horizontal path:

$x(t) = -t^2 + 4t - 2$

where $x(t)$ represents the position in meters, and $t$ is time in seconds. The problem asks for the distance traveled by the object during the time interval $0 \leq t \leq 5$.

Solution Outline

1. Velocity function: The velocity is the derivative of the position function: $v(t) = \frac{dx(t)}{dt} = -2t + 4$

2. Identify when velocity changes sign: The object’s direction changes when $v(t) = 0$, which occurs at: $-2t + 4 = 0 \implies t = 2 \text{ seconds}$ So, the object changes direction at $t = 2$.

3. Calculate positions at critical points: We need to evaluate the position function $x(t)$ at $t = 0$, $t = 2$, and $t = 5$.

   • $x(0) = -(0)^2 + 4(0) - 2 = -2$
   • $x(2) = -(2)^2 + 4(2) - 2 = 2$
   • $x(5) = -(5)^2 + 4(5) - 2 = -7$

4. Total distance traveled: Since the object changes direction at $t = 2$, we break the distance into two parts:

   • Distance from $t = 0$ to $t = 2$: $|x(2) - x(0)| = |2 - (-2)| = 4$
   • Distance from $t = 2$ to $t = 5$: $|x(5) - x(2)| = |-7 - 2| = 9$

   The total distance traveled is: $4 + 9 = 13 \text{ meters}$

Final Answer:

The object travels 13 meters in total during the interval $0 \leq t \leq 5$.

Would you like more details, or do you have any questions?

Here are five follow-up questions related to this problem:

1. How would the total distance change if the time interval were $0 \leq t \leq 6$?
2. What is the object's velocity at $t = 1$ and $t = 4$?
3. How can we find the total displacement instead of the total distance?
4. What would happen to the object's motion if the equation of $x(t)$ were changed to $x(t) = -t^2 + 6t - 5$?
5. How is the concept of total distance traveled different from displacement in physics?

Tip: The total distance is always the sum of the absolute values of changes in position, even if the object changes direction.