https://file.aiquickdraw.com/mathbot/file/1728859767159hm3irank.jpg

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

### Step 1: Find the velocity function

Velocity is the derivative of the position function \( s(t) \) with respect to time \( t \):

\[
v(t) = \frac{d}{dt} \left( \sqrt{13 + 6t} \right)
\]

Using the chain rule, we differentiate:

\[
v(t) = \frac{1}{2\sqrt{13 + 6t}} \cdot \frac{d}{dt}(13 + 6t) = \frac{1}{2\sqrt{13 + 6t}} \cdot 6
\]

This simplifies to:

\[
v(t) = \frac{6}{2\sqrt{13 + 6t}} = \frac{3}{\sqrt{13 + 6t}}
\]

Now, we substitute \( t = 2 \) into the velocity function:

\[
v(2) = \frac{3}{\sqrt{13 + 6(2)}} = \frac{3}{\sqrt{13 + 12}} = \frac{3}{\sqrt{25}} = \frac{3}{5} = 0.6 \, \text{m/s}
\]

### Step 2: Find the acceleration function

Acceleration is the derivative of the velocity function \( v(t) \) with respect to time \( t \):

\[
a(t) = \frac{d}{dt} \left( \frac{3}{\sqrt{13 + 6t}} \right)
\]

We apply the chain rule again. First, we rewrite the velocity function as \( v(t) = 3(13 + 6t)^{-\frac{1}{2}} \). Differentiating:

\[
a(t) = 3 \cdot \left( -\frac{1}{2} \right) \cdot (13 + 6t)^{-\frac{3}{2}} \cdot 6 = - \frac{9}{(13 + 6t)^{\frac{3}{2}}}
\]

Now, substitute \( t = 2 \) into the acceleration function:

\[
a(2) = - \frac{9}{(13 + 12)^{\frac{3}{2}}} = - \frac{9}{25^{\frac{3}{2}}} = - \frac{9}{125} = -0.072 \, \text{m/s}^2
\]

It seems there is a small discrepancy between my result and the one shown in the image, which lists acceleration as \( -0.144 \, \text{m/s}^2 \). This could be due to a different interpretation of the problem or a typo. Let me know if you'd like to investigate further.

Would you like more details or have any questions on this?

---

Here are 5 related questions:
1. How would the velocity change if the function was \( s = \sqrt{13 + 4t} \)?
2. Can you find the particle's velocity at \( t = 3 \) seconds?
3. What happens to acceleration if the position function involves higher powers of \( t \)?
4. How can you find the total distance traveled between \( t = 0 \) and \( t = 2 \)?
5. How does the shape of the position function affect the particle's motion?

**Tip**: Always remember to apply the chain rule carefully, especially when dealing with square roots or other composite functions.

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 seconds.

Find the velocity and acceleration of a particle moving along a coordinate line, with position s = √(13 + 6t). This problem demonstrates the use of differentiation and the chain rule in calculus to find velocity and acceleration at t = 2 seconds. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation