The equation of motion of a particle is 

*s* = *t*3 − 12*t*,

 where 

*s*

 is measured in meters and *t* is in seconds. (Assume 

*t* ≥ 0.)

(a)

Find the velocity and acceleration as functions of *t*.

*v*(*t*) = 
 
*a*(*t*) = 
 
(b)

Find the acceleration, in m/s2, after 4 seconds.

 m/s2

(c)

Find the acceleration, in m/s2, when the velocity is 0.

 m/s2

We are given the equation of motion for a particle as:

\[
s(t) = t^3 - 12t
\]

where \( s \) is the position in meters and \( t \) is the time in seconds.

### Part (a): Finding velocity and acceleration as functions of time

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

\[
v(t) = \frac{ds(t)}{dt}
\]

\[
v(t) = 3t^2 - 12
\]

2. **Acceleration (\( a(t) \))** is the derivative of the velocity function with respect to time:

\[
a(t) = \frac{dv(t)}{dt}
\]

\[
a(t) = 6t
\]

Thus, the velocity and acceleration as functions of \( t \) are:

\[
v(t) = 3t^2 - 12
\]

\[
a(t) = 6t
\]

### Part (b): Finding the acceleration after 4 seconds

We use the acceleration function \( a(t) = 6t \) and plug in \( t = 4 \):

\[
a(4) = 6(4) = 24 \, \text{m/s}^2
\]

Thus, the acceleration after 4 seconds is \( 24 \, \text{m/s}^2 \).

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

To find when the velocity is 0, we set the velocity equation \( v(t) = 3t^2 - 12 \) equal to 0 and solve for \( t \):

\[
3t^2 - 12 = 0
\]

\[
t^2 = 4
\]

\[
t = 2 \, \text{seconds} \quad (\text{since } t \geq 0)
\]

Now, we find the acceleration at \( t = 2 \):

\[
a(2) = 6(2) = 12 \, \text{m/s}^2
\]

Thus, the acceleration when the velocity is 0 is \( 12 \, \text{m/s}^2 \).

### Summary:
- **(a)** \( v(t) = 3t^2 - 12 \) and \( a(t) = 6t \)
- **(b)** The acceleration after 4 seconds is \( 24 \, \text{m/s}^2 \)
- **(c)** The acceleration when the velocity is 0 is \( 12 \, \text{m/s}^2 \)

Let me know if you would like further details or clarifications!

Here are some follow-up questions:

1. How do you interpret the relationship between velocity and acceleration?
2. What is the significance of negative velocity in this context?
3. How would you find the time when the particle comes to rest?
4. How does the motion change after the particle comes to rest?
5. How can you find the maximum or minimum displacement using calculus?

**Tip:** To find maximum or minimum values in a motion problem, take the derivative (velocity) and set it equal to 0, then check the second derivative (acceleration) to determine if it's a maximum or minimum!

The equation of motion of a particle is s = t^3 − 12t, where s is measured in meters and t is in seconds. (Assume t ≥ 0.) (a) Find the velocity and acceleration as functions of t. (b) Find the acceleration, in m/s^2, after 4 seconds. (c) Find the acceleration, in m/s^2, when the velocity is 0.

Discover how to calculate velocity and acceleration from the equation of motion s = t^3 − 12t, using calculus-based differentiation. This problem provides a step-by-step solution involving velocity and acceleration formulas. Ideal for high school students in Grades 11-12.

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