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.