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The image you uploaded contains a calculus problem related to finding the displacement of a particle. The problem is as follows:

Given the velocity of a particle as \( v(t) = 5t^2 \) (where \( t \) is time), the question asks how far the particle has traveled from \( t = 1 \) to \( t = 3 \).

### Solution:
To find the distance traveled by the particle, you need to calculate the definite integral of the velocity function from \( t = 1 \) to \( t = 3 \):

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

Now, calculate the integral:

\[
\int 5t^2 \, dt = \frac{5t^3}{3} + C
\]

Evaluate the definite integral from \( t = 1 \) to \( t = 3 \):

\[
\left[ \frac{5t^3}{3} \right]_{1}^{3} = \frac{5(3)^3}{3} - \frac{5(1)^3}{3}
\]

Simplify the expression:

\[
= \frac{5(27)}{3} - \frac{5(1)}{3} = \frac{135}{3} - \frac{5}{3} = 45 - \frac{5}{3} = \frac{135}{3} - \frac{5}{3} = \frac{130}{3} = 43.33
\]

The exact distance traveled by the particle is \( 43.33 \) units.

### Answer:
The correct answer from the options given is likely meant to be \( \frac{40}{3} \) units, but the calculation shows that \( 43.33 \) units (approximately \( 43 \frac{1}{3} \)) is the correct distance.

### Would you like more details on this solution, or do you have any questions?

Here are 5 related questions that expand on this information:
1. How would the distance change if the velocity function was \( v(t) = 5t^3 \) instead of \( 5t^2 \)?
2. How can you determine the displacement of the particle if the velocity function had negative values?
3. What is the significance of the constant of integration \( C \) when evaluating indefinite integrals?
4. How would the problem change if you were asked to find the total distance traveled instead of the net displacement?
5. How would you interpret the result if the velocity function was given as \( v(t) = -5t^2 \)?

**Tip:** Remember, when integrating a function to find displacement or distance, be mindful of the limits of integration and ensure the velocity function is integrated properly over the specified interval.

Given that v(t) = 5t^2 is the velocity of a particle at time t, how far has the particle traveled from t = 1 to t = 3?

Learn how to calculate the distance traveled by a particle given the velocity function v(t) = 5t^2, using definite integration from t = 1 to t = 3. This problem involves key calculus concepts such as integration and the Fundamental Theorem of Calculus, suitable for students in Grades 11-12.

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