Find the components of velocity and acceleration in the direction of a given unit vector at t=1.

The velocity vector is the first derivative of the position vector with respect to time.

$$\vec{v}(t) = \frac{dx}{dt} \mathbf{i} + \frac{dy}{dt} \mathbf{j} + \frac{dz}{dt} \mathbf{k}$$ $$\frac{dx}{dt} = \frac{d}{dt}(2t + 3) = 2$$ $$\frac{dy}{dt} = \frac{d}{dt}(t^2 + 3t) = 2t + 3$$ $$\frac{dz}{dt} = \frac{d}{dt}(t^3 + 2t^2) = 3t^2 + 4t$$

So, the velocity vector is:

$$\vec{v}(t) = 2\mathbf{i} + (2t + 3)\mathbf{j} + (3t^2 + 4t)\mathbf{k}$$

Substitute $t=1$ into the velocity vector:

$$\vec{v}(1) = 2\mathbf{i} + (2(1) + 3)\mathbf{j} + (3(1)^2 + 4(1))\mathbf{k}$$ $$\vec{v}(1) = 2\mathbf{i} + (2 + 3)\mathbf{j} + (3 + 4)\mathbf{k}$$ $$\vec{v}(1) = 2\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}$$

The acceleration vector is the first derivative of the velocity vector (or the second derivative of the position vector) with respect to time.

$$\vec{a}(t) = \frac{d^2x}{dt^2} \mathbf{i} + \frac{d^2y}{dt^2} \mathbf{j} + \frac{d^2z}{dt^2} \mathbf{k}$$ $$\frac{d^2x}{dt^2} = \frac{d}{dt}(2) = 0$$ $$\frac{d^2y}{dt^2} = \frac{d}{dt}(2t + 3) = 2$$ $$\frac{d^2z}{dt^2} = \frac{d}{dt}(3t^2 + 4t) = 6t + 4$$

So, the acceleration vector is:

$$\vec{a}(t) = 0\mathbf{i} + 2\mathbf{j} + (6t + 4)\mathbf{k}$$

Substitute $t=1$ into the acceleration vector:

$$\vec{a}(1) = 0\mathbf{i} + 2\mathbf{j} + (6(1) + 4)\mathbf{k}$$ $$\vec{a}(1) = 0\mathbf{i} + 2\mathbf{j} + (6 + 4)\mathbf{k}$$ $$\vec{a}(1) = 0\mathbf{i} + 2\mathbf{j} + 10\mathbf{k}$$

The given direction vector is $\vec{d} = 2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}$.
The magnitude of $\vec{d}$ is:

$$|\vec{d}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$$

The unit vector $\hat{u}$ in this direction is:

$$\hat{u} = \frac{\vec{d}}{|\vec{d}|} = \frac{1}{\sqrt{29}}(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})$$

The component of velocity $\vec{v}(1)$ in the direction of $\hat{u}$ is given by the dot product $\vec{v}(1) \cdot \hat{u}$:

$$\text{Component of velocity} = (2\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}) \cdot \frac{1}{\sqrt{29}}(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})$$ $$= \frac{1}{\sqrt{29}}((2)(2) + (5)(3) + (7)(4))$$ $$= \frac{1}{\sqrt{29}}(4 + 15 + 28)$$ $$= \frac{47}{\sqrt{29}}$$

The component of acceleration $\vec{a}(1)$ in the direction of $\hat{u}$ is given by the dot product $\vec{a}(1) \cdot \hat{u}$:

$$\text{Component of acceleration} = (0\mathbf{i} + 2\mathbf{j} + 10\mathbf{k}) \cdot \frac{1}{\sqrt{29}}(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})$$ $$= \frac{1}{\sqrt{29}}((0)(2) + (2)(3) + (10)(4))$$ $$= \frac{1}{\sqrt{29}}(0 + 6 + 40)$$ $$= \frac{46}{\sqrt{29}}$$

The component of velocity in the direction of $2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}$ at $t=1$ is $\boxed{\frac{47}{\sqrt{29}}}$.
The component of acceleration in the direction of $2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}$ at $t=1$ is $\boxed{\frac{46}{\sqrt{29}}}$.