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To solve for the total distance traveled by an object moving with velocity \( v = t - 5 \) between \( t = 0 \) and \( t = 6 \), we need to integrate the absolute value of the velocity function over this interval.

The velocity function is \( v(t) = t - 5 \). The object changes direction when the velocity is zero, which happens when \( t - 5 = 0 \) or \( t = 5 \).

We will split the integral into two parts: from \( t = 0 \) to \( t = 5 \), and from \( t = 5 \) to \( t = 6 \).

1. **From \( t = 0 \) to \( t = 5 \):**
   The velocity \( v(t) = t - 5 \) is negative because \( t - 5 < 0 \). Therefore, the speed (absolute value of velocity) is \( |t - 5| = 5 - t \).

   \[
   \text{Distance} = \int_{0}^{5} (5 - t) \, dt
   \]

   \[
   \int_{0}^{5} (5 - t) \, dt = \left[ 5t - \frac{t^2}{2} \right]_{0}^{5}
   \]

   \[
   = \left( 5(5) - \frac{5^2}{2} \right) - \left( 5(0) - \frac{0^2}{2} \right)
   \]

   \[
   = \left( 25 - \frac{25}{2} \right) - 0 = 25 - 12.5 = 12.5
   \]

2. **From \( t = 5 \) to \( t = 6 \):**
   The velocity \( v(t) = t - 5 \) is positive because \( t - 5 > 0 \). Therefore, the speed is \( |t - 5| = t - 5 \).

   \[
   \text{Distance} = \int_{5}^{6} (t - 5) \, dt
   \]

   \[
   \int_{5}^{6} (t - 5) \, dt = \left[ \frac{(t - 5)^2}{2} \right]_{5}^{6}
   \]

   \[
   = \left( \frac{(6 - 5)^2}{2} \right) - \left( \frac{(5 - 5)^2}{2} \right)
   \]

   \[
   = \left( \frac{1^2}{2} \right) - 0 = \frac{1}{2} = 0.5
   \]

Finally, adding these two distances gives the total distance traveled:

\[
12.5 + 0.5 = 13
\]

Therefore, the total distance the object travels between \( t = 0 \) and \( t = 6 \) is **13** units, so the correct answer is **B**.

Would you like more details on any specific part of this solution? Here are five related questions to consider:

1. How would the distance calculation change if the velocity function were different, such as \( v = t^2 - 5 \)?
2. Can you explain how to determine when an object changes direction given a different velocity function?
3. What is the significance of integrating the absolute value of the velocity function in distance calculations?
4. How can we use the concept of definite integrals to find the displacement instead of the total distance?
5. What would the total distance be if the velocity function were \( v = t - 2 \) over the same interval?

**Tip:** When dealing with integrals of absolute value functions, always find where the function changes sign to properly split the integral.

An object moves along a horizontal axis with velocity v = t - 5. What is the total distance the object travels between t = 0 and t = 6?

In this calculus problem, we find the total distance traveled by an object moving with velocity v = t - 5 between t = 0 and t = 6. The solution involves splitting the integral to account for changes in direction and calculating the definite integral of the absolute value of the velocity.

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