To find the velocity and speed of a particle at a specific time $t = 2$, we need to first differentiate the position function $s = f(t)$ to obtain the velocity function, and then evaluate it at $t = 2$.

Given the position function:

$f(t) = 12 + \frac{15}{t + 1}$

Step 1: Find the Velocity Function

The velocity $v(t)$ is the derivative of the position function $f(t)$ with respect to time $t$:

$v(t) = \frac{d}{dt} \left( 12 + \frac{15}{t + 1} \right)$

Let's differentiate this function.

$v(t) = 0 + 15 \cdot \frac{d}{dt} \left( \frac{1}{t + 1} \right)$

Using the derivative rule $\frac{d}{dt} \left( \frac{1}{t + 1} \right) = -\frac{1}{(t + 1)^2}$:

$v(t) = 15 \cdot \left( -\frac{1}{(t + 1)^2} \right) = -\frac{15}{(t + 1)^2}$

Step 2: Evaluate the Velocity at $t = 2$

Substitute $t = 2$ into the velocity function:

$v(2) = -\frac{15}{(2 + 1)^2} = -\frac{15}{3^2} = -\frac{15}{9} = -\frac{5}{3} \, \text{m/s}$

Step 3: Find the Speed

The speed is the absolute value of the velocity:

$\text{Speed} = |v(2)| = \left| -\frac{5}{3} \right| = \frac{5}{3} \, \text{m/s}$

Final Answers

• Velocity at $t = 2$: $-\frac{5}{3} \, \text{m/s}$
• Speed at $t = 2$: $\frac{5}{3} \, \text{m/s}$

Would you like further details or have any questions?

Related Questions:

1. What is the acceleration of the particle at $t = 2$?
2. How does the velocity change as $t$ increases?
3. What is the position of the particle at $t = 2$?
4. How would you find the maximum or minimum velocity of the particle?
5. How does the speed compare to velocity in general motion?

Tip: Remember, speed is always positive because it is the magnitude of velocity!