The velocity of a particle is the time derivative of its position vector.

Given the position vector of the particle:

$\mathbf{R_1}(t) = (2t + 1)\hat{i} + (t + 3)\hat{j} - (2t - 3)\hat{k}$

We can compute the velocity vector $\mathbf{v}(t)$ by differentiating $\mathbf{R_1}(t)$ with respect to time $t$:

$\mathbf{v}(t) = \frac{d\mathbf{R_1}(t)}{dt}$

Let's differentiate each component of $\mathbf{R_1}(t)$ with respect to $t$:

1. The $\hat{i}$-component: $2t + 1$ becomes: $\frac{d}{dt}(2t + 1) = 2$

2. The $\hat{j}$-component: $t + 3$ becomes: $\frac{d}{dt}(t + 3) = 1$

3. The $\hat{k}$-component: $-2t + 3$ becomes: $\frac{d}{dt}(-2t + 3) = -2$

Thus, the velocity vector is:

$\mathbf{v}(t) = 2\hat{i} + 1\hat{j} - 2\hat{k}$

Final Answer:

The velocity of the particle is:

$\mathbf{v}(t) = 2\hat{i} + 1\hat{j} - 2\hat{k} \, \text{cm/s}$

Would you like further details or clarifications on this? Here are some related questions for you to explore:

1. How do you interpret the velocity vector in terms of motion in 3D space?
2. What is the acceleration of the particle if its velocity is constant?
3. How would you find the speed of the particle from the velocity vector?
4. How does the position vector differ from the velocity vector in describing particle motion?
5. Can you explain how the time derivatives of each component of the position vector lead to the velocity components?

Tip: The velocity vector gives you both the magnitude and direction of the particle's motion. When the components are constants (like in this case), the velocity is uniform, meaning the particle moves with constant speed in a straight line.