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The given equation of motion for a particle is:

\[
s(t) = \frac{1}{3}t^3 - 4t^2 + 16t + 6
\]

### (a) Finding the velocity and acceleration as functions of \( t \)

1. **Velocity** \( v(t) \) is the first derivative of \( s(t) \) with respect to time \( t \):

\[
v(t) = \frac{ds}{dt} = \frac{d}{dt} \left( \frac{1}{3}t^3 - 4t^2 + 16t + 6 \right)
\]

Calculating the derivative:

\[
v(t) = t^2 - 8t + 16
\]

2. **Acceleration** \( a(t) \) is the derivative of velocity, or the second derivative of position \( s(t) \):

\[
a(t) = \frac{dv}{dt} = \frac{d}{dt} \left( t^2 - 8t + 16 \right)
\]

Calculating the derivative:

\[
a(t) = 2t - 8
\]

Thus, the velocity and acceleration as functions of \( t \) are:
\[
v(t) = t^2 - 8t + 16
\]
\[
a(t) = 2t - 8
\]

### (b) Finding the acceleration after 1 second

Substitute \( t = 1 \) into the acceleration function:

\[
a(1) = 2(1) - 8 = 2 - 8 = -6 \, \text{m/s}^2
\]

So, the acceleration after 1 second is:

\[
a(1) = -6 \, \text{m/s}^2
\]

### (c) Finding the acceleration when the velocity is 0

Set the velocity function equal to 0 to find \( t \):

\[
v(t) = t^2 - 8t + 16 = 0
\]

Solve the quadratic equation:

\[
t^2 - 8t + 16 = (t - 4)^2 = 0
\]

This gives \( t = 4 \) seconds.

Now, substitute \( t = 4 \) into the acceleration function:

\[
a(4) = 2(4) - 8 = 8 - 8 = 0 \, \text{m/s}^2
\]

Thus, the acceleration when the velocity is 0 is:

\[
a(4) = 0 \, \text{m/s}^2
\]

---

Would you like more details or have any questions? Here are five related questions you might find interesting:

1. How would you find the position at a given time \( t \)?
2. What is the maximum velocity of the particle?
3. At what times is the particle's acceleration zero?
4. How would you calculate the jerk (the derivative of acceleration) for this particle?
5. How would the motion change if a constant external force were applied?

**Tip:** Always check the units when performing calculations in physics to ensure consistency, especially when mixing different physical quantities like position, velocity, and acceleration.

Suppose that the equation of motion for a particle (where s is in meters and t in seconds) is s = (1/3)t^3 - 4t^2 + 16t + 6.
(a) Find the velocity and acceleration as functions of t.
(b) Find the acceleration after 1 second.
(c) Find the acceleration at the instant when the velocity is 0.

Learn how to find the velocity and acceleration from the given equation of motion: s = (1/3)t^3 - 4t^2 + 16t + 6. Detailed step-by-step solution using differentiation to solve for velocity, acceleration at different points, and when the velocity is zero. Suitable for Grades 11-12.

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