https://file.aiquickdraw.com/mathbot/file/17288596256239ywv4rdw.jpg

The given problem states that the position of a particle moving along a coordinate line is \( s = \sqrt{13 + 6t} \), with \( s \) in meters and \( t \) in seconds. We are asked to find the velocity and acceleration of the particle at \( t = 2 \) seconds.

### Step 1: Find the velocity
The velocity is the time derivative of the position function \( s(t) \).

Given:
\[
s(t) = \sqrt{13 + 6t}
\]

Differentiate \( s(t) \) with respect to \( t \):
\[
v(t) = \frac{ds}{dt} = \frac{d}{dt} \left( \sqrt{13 + 6t} \right)
\]

Using the chain rule:
\[
v(t) = \frac{1}{2\sqrt{13 + 6t}} \cdot 6 = \frac{3}{\sqrt{13 + 6t}}
\]

Now, substitute \( t = 2 \) into the velocity equation:
\[
v(2) = \frac{3}{\sqrt{13 + 6(2)}} = \frac{3}{\sqrt{13 + 12}} = \frac{3}{\sqrt{25}} = \frac{3}{5}
\]

So, the velocity at \( t = 2 \) seconds is:
\[
v(2) = 0.6 \, \text{m/s}
\]

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

From the velocity function:
\[
v(t) = \frac{3}{\sqrt{13 + 6t}}
\]

Differentiate \( v(t) \) with respect to \( t \):
\[
a(t) = \frac{d}{dt} \left( \frac{3}{\sqrt{13 + 6t}} \right)
\]

Again using the chain rule:
\[
a(t) = -\frac{3}{2(13 + 6t)^{3/2}} \cdot 6 = -\frac{18}{(13 + 6t)^{3/2}}
\]

Substitute \( t = 2 \) into the acceleration equation:
\[
a(2) = -\frac{18}{(13 + 6(2))^{3/2}} = -\frac{18}{(13 + 12)^{3/2}} = -\frac{18}{25^{3/2}} = -\frac{18}{125}
\]

So, the acceleration at \( t = 2 \) seconds is:
\[
a(2) = -0.144 \, \text{m/s}^2
\]

### Final Answer:
- Velocity at \( t = 2 \) sec: \( 0.6 \, \text{m/s} \)
- Acceleration at \( t = 2 \) sec: \( -0.144 \, \text{m/s}^2 \)

Would you like more details or have any further questions?

Here are 5 follow-up questions to extend your understanding:
1. How would the velocity change if the position function was different?
2. What is the significance of the negative sign in the acceleration result?
3. Can you find the time when the velocity is zero?
4. How does the velocity behave as \( t \) approaches infinity?
5. What would happen to the velocity if we modified the constant in the square root?

**Tip:** The chain rule is essential when differentiating composite functions like square roots. Always remember to apply it carefully!

The position of a particle moving along a coordinate line is s = √(13 + 6t), with s in meters and t in seconds. Find the particle's velocity and acceleration at t = 2 sec.

Calculate the velocity and acceleration of a particle given the position function s = √(13 + 6t). This problem involves applying calculus concepts such as differentiation, the chain rule, and finding instantaneous rates of change. Suitable for students in Grades 11-12 or introductory college calculus courses.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation