The displacement $s(t)$ of a particle is given by the equation:

$s(t) = 6t^2$

The velocity $v(t)$ of the particle is the derivative of the displacement function $s(t)$ with respect to time $t$. So, we find $v(t)$ by differentiating $s(t)$.

$v(t) = \frac{ds}{dt} = \frac{d}{dt} \left( 6t^2 \right) = 12t$

Now, we can find the velocity at different times.

1. At $t = a$: $v(a) = 12a$

2. At $t = 1$: $v(1) = 12 \times 1 = 12 \ \text{m/s}$

3. At $t = 2$: $v(2) = 12 \times 2 = 24 \ \text{m/s}$

4. At $t = 3$: $v(3) = 12 \times 3 = 36 \ \text{m/s}$

Summary of velocities:

• $v(a) = 12a$
• $v(1) = 12 \ \text{m/s}$
• $v(2) = 24 \ \text{m/s}$
• $v(3) = 36 \ \text{m/s}$

Would you like more details or have any questions?

---

Relative Questions:

1. How do we interpret the derivative of a function in the context of motion?
2. What is the significance of the constant coefficient in the velocity equation?
3. How would you find the acceleration of the particle at a given time?
4. What happens to the velocity as $t$ increases?
5. Can we use the same process to find displacement from velocity?

Tip: Derivatives are essential in physics, as they often represent rates of change, such as velocity from displacement or acceleration from velocity.