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The image you've uploaded contains a problem about the motion of a particle along a straight line, where its position as a function of time is given. The tasks are to find the velocity and acceleration functions and evaluate them at a specific time.

Here’s the problem:

Given the position function \( s(t) = t^3 - 12t^2 + 21t \) where \( s \) is in meters and \( t \) is in seconds:

1. **Find the velocity function \( v(t) \) as a function of \( t \).**
2. **Find the acceleration function \( a(t) \) as a function of \( t \).**
3. **Evaluate the velocity at \( t = 2 \) seconds.**
4. **Evaluate the acceleration at \( t = 2 \) seconds.**

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

\[
v(t) = \frac{ds(t)}{dt} = \frac{d}{dt}\left( t^3 - 12t^2 + 21t \right)
\]

Let's compute this derivative:

\[
v(t) = 3t^2 - 24t + 21
\]

### Step 2: Find the Acceleration Function \( a(t) \)
The acceleration is the derivative of the velocity function with respect to time \( t \):

\[
a(t) = \frac{dv(t)}{dt} = \frac{d}{dt}\left( 3t^2 - 24t + 21 \right)
\]

Let's compute this derivative:

\[
a(t) = 6t - 24
\]

### Step 3: Evaluate the Velocity at \( t = 2 \) seconds
Now substitute \( t = 2 \) into the velocity function:

\[
v(2) = 3(2)^2 - 24(2) + 21 = 3(4) - 48 + 21 = 12 - 48 + 21 = -15 \, \text{m/s}
\]

### Step 4: Evaluate the Acceleration at \( t = 2 \) seconds
Now substitute \( t = 2 \) into the acceleration function:

\[
a(2) = 6(2) - 24 = 12 - 24 = -12 \, \text{m/s}^2
\]

### Summary:
- The velocity function is \( v(t) = 3t^2 - 24t + 21 \).
- The acceleration function is \( a(t) = 6t - 24 \).
- The velocity at \( t = 2 \) seconds is \( v(2) = -15 \, \text{m/s} \).
- The acceleration at \( t = 2 \) seconds is \( a(2) = -12 \, \text{m/s}^2 \).

Would you like any additional details or have any questions?

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### Follow-up Questions:
1. What is the significance of the sign of velocity and acceleration in this context?
2. How would you interpret the particle’s motion if the acceleration were positive at \( t = 2 \) seconds?
3. How would the velocity function change if an additional constant term were added to the position function \( s(t) \)?
4. Can you find the time when the particle is at rest (i.e., when the velocity is zero)?
5. What does the position function look like graphically, and how do velocity and acceleration relate to its curvature?

### Tip:
Remember that the velocity function represents the rate of change of position, while the acceleration function represents the rate of change of velocity. This can help you understand the particle's motion more intuitively.

A particle is moving along a straight line, and its position with respect to a reference point is given by the equation s(t) = t^3 - 12t^2 + 21t, where s is in meters and t is in seconds. The tasks are to: 1) Find the velocity function v(t) as a function of time. 2) Find the acceleration function a(t) as a function of time. 3) Evaluate the velocity at t = 2 seconds. 4) Evaluate the acceleration at t = 2 seconds.

Learn how to find the velocity and acceleration of a particle moving along a straight line using the position function s(t) = t^3 - 12t^2 + 21t. This problem includes step-by-step differentiation to obtain the velocity and acceleration functions and evaluates them at t = 2 seconds. Suitable for high school students.

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