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To determine when the particle is at rest, we need to find when its velocity is zero. The velocity \( v(t) \) is the derivative of the position function \( x(t) \).

The position function is:
\[
x(t) = 2t^3 - 21t^2 + 72t - 53
\]

### Step 1: Differentiate \( x(t) \) to find \( v(t) \)
\[
v(t) = \frac{dx}{dt} = 6t^2 - 42t + 72
\]

### Step 2: Set \( v(t) = 0 \) to find the times when the particle is at rest
\[
6t^2 - 42t + 72 = 0
\]

Divide through by 6:
\[
t^2 - 7t + 12 = 0
\]

Now, factor the quadratic:
\[
(t - 3)(t - 4) = 0
\]

So, \( t = 3 \) and \( t = 4 \).

### Conclusion
The particle is at rest at \( t = 3 \) and \( t = 4 \). Thus, the correct answer is:
\[
\text{(E) } t = 3 \text{ and } t = 4
\]

Would you like more details, or do you have any questions?

### Related Questions
1. How would we interpret the particle’s motion if the velocity had no real roots?
2. What is the significance of the sign of \( v(t) \) in determining the direction of the particle?
3. How can we find the particle’s acceleration at specific points?
4. If \( t \) represented time in seconds, what units would \( x(t) \) have?
5. Can the particle come to rest at more than two points?

### Tip
For particle motion problems, remember that setting the derivative of the position function equal to zero finds the times when the particle is at rest.

A particle moves along the x-axis so that at time t ≥ 0 its position is given by x(t) = 2t^3 - 21t^2 + 72t - 53. At what time t is the particle at rest?

Learn how to find when a particle is at rest by solving the derivative of its position function. This problem involves the derivative of a cubic function and solving a quadratic equation. Suitable for students in Grades 11-12.

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