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!