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The problem asks to determine the velocity and acceleration of a particle at \( t = 4 \) seconds, given the position function \( s(t) = \sqrt{1 + 6t} \), where \( s \) is in meters and \( t \) is in seconds.

### Step 1: Find the velocity
Velocity is the first derivative of the position function with respect to time, \( v(t) = \frac{ds}{dt} \).

Given:
\[
s(t) = \sqrt{1 + 6t}
\]

Apply the chain rule to differentiate:

\[
v(t) = \frac{d}{dt} \left( \sqrt{1 + 6t} \right)
\]
Using the chain rule:

\[
v(t) = \frac{1}{2\sqrt{1 + 6t}} \cdot \frac{d}{dt}(1 + 6t)
\]
\[
v(t) = \frac{1}{2\sqrt{1 + 6t}} \cdot 6
\]
\[
v(t) = \frac{6}{2\sqrt{1 + 6t}} = \frac{3}{\sqrt{1 + 6t}}
\]

Now, calculate the velocity at \( t = 4 \) seconds:

\[
v(4) = \frac{3}{\sqrt{1 + 6(4)}} = \frac{3}{\sqrt{1 + 24}} = \frac{3}{\sqrt{25}} = \frac{3}{5}
\]

So, the velocity at \( t = 4 \) seconds is:
\[
v(4) = 0.6 \, \text{m/s}
\]

### Step 2: Find the acceleration
Acceleration is the second derivative of the position function with respect to time, \( a(t) = \frac{dv}{dt} \).

We already know that:
\[
v(t) = \frac{3}{\sqrt{1 + 6t}}
\]

Differentiate \( v(t) \) with respect to \( t \):

\[
a(t) = \frac{d}{dt} \left( \frac{3}{\sqrt{1 + 6t}} \right)
\]
Using the chain rule:

\[
a(t) = 3 \cdot \frac{-1}{2(1 + 6t)^{3/2}} \cdot 6
\]
\[
a(t) = \frac{-18}{2(1 + 6t)^{3/2}}
\]
\[
a(t) = \frac{-9}{(1 + 6t)^{3/2}}
\]

Now, calculate the acceleration at \( t = 4 \) seconds:

\[
a(4) = \frac{-9}{(1 + 6(4))^{3/2}} = \frac{-9}{(1 + 24)^{3/2}} = \frac{-9}{25^{3/2}} = \frac{-9}{125}
\]

So, the acceleration at \( t = 4 \) seconds is:
\[
a(4) = -0.072 \, \text{m/s}^2
\]

### Final Answers:
- Velocity at \( t = 4 \) seconds: \( v(4) = 0.6 \, \text{m/s} \)
- Acceleration at \( t = 4 \) seconds: \( a(4) = -0.072 \, \text{m/s}^2 \)

Would you like any more details or have further questions?

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Here are 5 related questions:
1. How does the velocity change with time for the given position function?
2. What is the significance of negative acceleration in this context?
3. How would the acceleration change if the position function was cubic instead of square root?
4. Can you find the time when the particle comes to rest?
5. How do you interpret the velocity and acceleration graphically for this motion?

**Tip**: The velocity is always the rate of change of position, and the acceleration is the rate of change of velocity—differentiating one function leads to valuable insights on the next!

La posición de una partícula que se desplaza a lo largo de una recta coordenada es s = √(1 + 6t), con s en metros y t en segundos. Determine la velocidad y la aceleración de la partícula cuando t = 4 segundos.

This problem explores how to calculate the velocity and acceleration of a particle moving along a coordinate line, given the position function s = √(1 + 6t). The detailed solution covers differentiation concepts, including the chain rule, and calculates the velocity and acceleration at t = 4 seconds.

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