91Ó°ÊÓ

Displacement in physics refers to the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. In this problem, the displacement is calculated by integrating the velocity function over the given interval of time, which is from 0 to 5 seconds.

The velocity function given is \( v(t) = t^2 - 6t + 8 \). To find the displacement, we integrate this function:

\[ \int_0^5 (t^2 - 6t + 8) \ dt \]

The solution involves finding the antiderivative of the velocity function and evaluating it from 0 to 5. The result of this calculation gives us the net change in position:

\[ \left[\frac{t^3}{3} - 3t^2 + 8t\right]_0^5 = \frac{10}{3} \text{ m} \]

This value represents how far and in what direction the object has moved from its starting point.

Distance traveled differs from displacement because it considers the total path covered, regardless of direction. It is a scalar quantity, meaning it has only magnitude. In cases where the object changes direction, as it does in this scenario, distance traveled can be greater than displacement.

The velocity equation \(v(t) = t^2 - 6t + 8\) changes sign, indicating direction changes. To find the total distance, we integrate the absolute value of the velocity function across each interval of motion.

• For \(0 \leq t < 2\) and \(4 < t \leq 5\), where the velocity is positive, we integrate normally.
• For \(2 < t < 4\), where the motion is negative, the velocity function is reversed.

Adding these integrals gives us the total distance traveled:

\[\frac{22}{3} \text{ m}\]

This distance is larger because it includes every part of the object's path, even as it moves backward.

Interval analysis involves breaking down the motion into different intervals based on where the velocity changes sign. This helps determine when the object is moving forward or backward.

Solving \( t^2 - 6t + 8 = 0\) gives \(t = 2\) and \(t = 4\), which splits the time into three intervals:

• \(0 \leq t < 2\)
• \(2 < t < 4\)
• \(4 < t \leq 5\)

Using test points in each interval allows us to check the velocity:

• For \(0 \leq t < 2\), \(v(1) > 0\), indicating positive motion.
• For \(2 < t < 4\), \(v(3) < 0\), indicating negative motion.
• For \(4 < t \leq 5\), \(v(5) > 0\), indicating positive motion again.

By understanding how velocity changes in these intervals, you can see the complete picture of the object's movement along its path.