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The position function of the particle is given as:

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

We are asked to find the velocity and position of the particle when its acceleration is zero.

### Step 1: Find the velocity function
The velocity \( v(t) \) is the derivative of the position function \( s(t) \) with respect to time \( t \):

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

Differentiating \( s(t) \):

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

### Step 2: Find the acceleration function
The acceleration \( a(t) \) is the derivative of the velocity function \( v(t) \):

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

Differentiating \( v(t) \):

\[
a(t) = 2t - 6
\]

### Step 3: Solve for when the acceleration is zero
We need to find when \( a(t) = 0 \):

\[
2t - 6 = 0
\]

Solving for \( t \):

\[
t = 3
\]

### Step 4: Find the velocity when \( t = 3 \)
Substitute \( t = 3 \) into the velocity function \( v(t) \):

\[
v(3) = 3^2 - 6(3) = 9 - 18 = -9
\]

Thus, the velocity when the acceleration is zero is \( v(3) = -9 \).

### Step 5: Find the position when \( t = 3 \)
Substitute \( t = 3 \) into the position function \( s(t) \):

\[
s(3) = \frac{3^3}{3} - 3(3^2) + 4 = \frac{27}{3} - 3(9) + 4 = 9 - 27 + 4 = -14
\]

Thus, the position when the acceleration is zero is \( s(3) = -14 \).

### Final Answer:
- The velocity of the particle when the acceleration is zero is \( v(3) = -9 \).
- The position of the particle when the acceleration is zero is \( s(3) = -14 \).

Would you like further details or have any questions?

#### 5 Related Questions:
1. How do we interpret negative velocity in this context?
2. What does it mean physically for the acceleration to be zero at a specific time?
3. How would the graph of the position function \( s(t) \) look?
4. What if the position function had more terms, like \( t^4 \)? How would the process change?
5. Can we use these techniques to find the maximum or minimum position of the particle?

#### Tip:
Always check the units of each physical quantity (position, velocity, and acceleration) to ensure consistency when solving real-world problems.

If the position of a particle is s(t) = t^3/3 - 3t^2 + 4, find the velocity and position of the particle when its acceleration is zero.

This problem focuses on finding the velocity and position of a particle given its position function s(t) = t^3/3 - 3t^2 + 4, when the acceleration is zero. It involves key calculus concepts such as derivatives, velocity, and acceleration. Suitable for students learning calculus in high school or college.

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