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The velocity \( v(t) \) of the object is found by taking the derivative of the position function \( s(t) \). The velocity function is:

\[
v(t) = t^2 - 7t + 6
\]

To find when the velocity is zero, we solve:

\[
t^2 - 7t + 6 = 0
\]

The solutions are \( t = 1 \) second and \( t = 6 \) seconds.

Would you like more details or have any questions?

Here are 5 related questions to consider:
1. How do you interpret the physical meaning of the velocity being zero?
2. What is the object's acceleration at \( t = 1 \) and \( t = 6 \)?
3. How do you find the maximum or minimum position of the object?
4. What is the object's position at \( t = 1 \) and \( t = 6 \)?
5. Can you determine when the object is moving forward or backward?

**Tip**: The critical points where velocity is zero often indicate where an object changes direction or stops momentarily.

If the position function of an object is given by s(t) = (1/3)t^3 - (7/2)t^2 + 6t, where s(t) is measured in meters and t is in seconds, find the time(s) when the velocity of the object is zero.

Learn how to find the time(s) when the velocity of an object is zero by differentiating its position function s(t) = (1/3)t^3 - (7/2)t^2 + 6t. This problem involves using calculus concepts such as derivatives and solving quadratic equations.

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