https://file.aiquickdraw.com/mathbot/file/1727392108329060x3fta.jpg

The velocity function is given as \( v(t) = t^3 - 5t^2 \), and we need to find both the displacement and the distance traveled by the particle over the time interval \([-1, 7]\).

### 1. Displacement
Displacement is the net change in position, which can be calculated by integrating the velocity function over the given time interval.

\[
\text{Displacement} = \int_{-1}^{7} v(t) \, dt = \int_{-1}^{7} (t^3 - 5t^2) \, dt
\]

I'll calculate the displacement first, and then we can proceed with finding the distance traveled.

### 2. Distance Traveled
The distance traveled requires integrating the absolute value of the velocity function. This involves determining when \( v(t) = 0 \), which will tell us when the particle changes direction. We will integrate in separate intervals depending on the sign of \( v(t) \).

Let's solve both these steps.### Displacement:
The displacement, which is the net change in position, is:
\[
\text{Displacement} = \frac{80}{3} \text{ meters} \approx 26.66 \text{ meters}
\]
This matches the value provided in the image.

### Distance Traveled:
To find the total distance traveled, we integrate the absolute value of the velocity over the intervals determined by the points where \( v(t) = 0 \). These points are approximately \( t = 0 \) and \( t = 5 \). The integral has to be computed over the segments \([-1, 0]\), \([0, 5]\), and \([5, 7]\), adjusting for the sign changes in the velocity function.

Let me finalize the distance calculation for these intervals.It seems the expression for the distance traveled involves integrating piecewise absolute values of the velocity function over the intervals \([0, 5]\) and \([5, 7]\). To simplify this and provide an exact numeric result, I'll compute these integrals next.It appears the integration of the absolute value over the second and third intervals is not immediately resolving to a simple expression. However, I will reattempt this step with an alternative approach to ensure we get the correct numerical value for the total distance traveled.

Let me handle the calculations explicitly for the remaining intervals.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, I can summarize the current process or provide further clarification based on the displacement.

Let me know how you’d like to proceed!

In the meantime, here are 5 related questions you could explore:
1. How does the displacement change if the velocity function has a different form?
2. What happens to the distance traveled if the time interval is increased?
3. How would you handle velocity functions with periodic behavior?
4. What are the key differences between displacement and distance in terms of motion?
5. Can you derive the position function from the given velocity function?

**Tip**: Always remember that displacement considers the direction of motion, while distance only considers the magnitude.

The velocity function (in meters per second) for a particle moving along a line is given by v(t) = t^3 - 5t^2. Find the displacement and the distance traveled by the particle during the time interval [-1,7].

Learn how to calculate both the displacement and the total distance traveled for a particle using the velocity function v(t) = t^3 - 5t^2 over the time interval [-1,7]. Detailed explanation with calculus integration and piecewise absolute value methods.

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