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To solve this problem, we need to determine the velocity and speed of the particle at \( t = 5 \) seconds. The position function of the particle is given by:

\[
f(t) = 18 + \frac{48}{t + 1}
\]

### Step 1: Find the Velocity

Velocity is the derivative of the position function \( f(t) \) with respect to time \( t \). We can find the velocity function by differentiating \( f(t) \).

Let's differentiate:

\[
f(t) = 18 + \frac{48}{t + 1}
\]

The derivative of \( f(t) \) is:

\[
f'(t) = 0 + \frac{d}{dt} \left(\frac{48}{t + 1}\right)
\]

Using the quotient rule:

\[
\frac{d}{dt} \left(\frac{48}{t + 1}\right) = 48 \cdot \frac{-1}{(t + 1)^2} = \frac{-48}{(t + 1)^2}
\]

So, the velocity function is:

\[
f'(t) = \frac{-48}{(t + 1)^2}
\]

### Step 2: Evaluate the Velocity at \( t = 5 \)

Substituting \( t = 5 \):

\[
f'(5) = \frac{-48}{(5 + 1)^2} = \frac{-48}{36} = -\frac{4}{3} \, \text{m/s}
\]

### Step 3: Find the Speed

Speed is the absolute value of velocity. Therefore:

\[
\text{Speed} = \left| -\frac{4}{3} \right| = \frac{4}{3} \, \text{m/s}
\]

### Final Answers:

- **Velocity at \( t = 5 \) is**: \(-\frac{4}{3} \, \text{m/s}\)
- **Speed at \( t = 5 \) is**: \(\frac{4}{3} \, \text{m/s}\)

Would you like more details on the differentiation process, or do you have any further questions?

### Related Questions:
1. How do you apply the quotient rule in differentiation?
2. What is the significance of velocity being negative?
3. How can you determine when the particle is at rest?
4. What does it mean if the speed is decreasing over time?
5. How would the problem change if the function had a different form?

### Tip:
Always check the units of your answer when solving physics problems; they help verify that your calculations are set up correctly.

A particle moves along a straight line with the equation of motion s = f(t), where s is measured in meters and t in seconds. The function given is f(t) = 18 + 48 / (t + 1). Find the velocity and speed (in m/s) when t = 5.

Learn how to find the velocity and speed of a particle moving along a straight line with equation f(t) = 18 + 48 / (t + 1). This detailed solution covers differentiation using the quotient rule to find velocity and absolute value for speed at t = 5 seconds. Suitable for students in Grades 10-12 studying calculus.

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