A particle has a velocity of 1i + 6j + 5k m/s. It has a constant tangential acceleration of 4 m/s², and a normal acceleration of 1i + 4j - 5k m/s². Find the x, y, and z components of its net acceleration.

The velocity vector is given as $\vec{v} = 1\hat{i} + 6\hat{j} + 5\hat{k}$ m/s.
First, calculate the magnitude of the velocity vector:

$$|\vec{v}| = \sqrt{1^2 + 6^2 + 5^2} = \sqrt{1 + 36 + 25} = \sqrt{62} \text{ m/s}$$

Next, find the unit vector in the direction of velocity:

$$\hat{u}_v = \frac{\vec{v}}{|\vec{v}|} = \frac{1\hat{i} + 6\hat{j} + 5\hat{k}}{\sqrt{62}}$$

The magnitude of the tangential acceleration is given as $a_t = 4$ m/s$^2$.
The tangential acceleration vector $\vec{a}_t$ is in the same direction as the velocity vector:

$$\vec{a}_t = a_t \hat{u}_v = 4 \left( \frac{1\hat{i} + 6\hat{j} + 5\hat{k}}{\sqrt{62}} \right) = \frac{4}{\sqrt{62}}\hat{i} + \frac{24}{\sqrt{62}}\hat{j} + \frac{20}{\sqrt{62}}\hat{k} \text{ m/s}^2$$

The normal acceleration vector is given as $\vec{a}_n = 1\hat{i} + 4\hat{j} - 5\hat{k}$ m/s$^2$.
The net acceleration $\vec{a}$ is the vector sum of the tangential and normal accelerations:

$$\vec{a} = \vec{a}_t + \vec{a}_n$$ $$\vec{a} = \left( \frac{4}{\sqrt{62}}\hat{i} + \frac{24}{\sqrt{62}}\hat{j} + \frac{20}{\sqrt{62}}\hat{k} \right) + (1\hat{i} + 4\hat{j} - 5\hat{k})$$

Combine the corresponding components:

$$\vec{a} = \left( 1 + \frac{4}{\sqrt{62}} \right)\hat{i} + \left( 4 + \frac{24}{\sqrt{62}} \right)\hat{j} + \left( -5 + \frac{20}{\sqrt{62}} \right)\hat{k}$$

From the net acceleration vector, the components are:

$$a_x = 1 + \frac{4}{\sqrt{62}}$$ $$a_y = 4 + \frac{24}{\sqrt{62}}$$ $$a_z = -5 + \frac{20}{\sqrt{62}}$$

Using $\sqrt{62} \approx 7.87400787$:

$$a_x \approx 1 + \frac{4}{7.87400787} \approx 1 + 0.50800000 \approx 1.5080 \text{ m/s}^2$$ $$a_y \approx 4 + \frac{24}{7.87400787} \approx 4 + 3.04800000 \approx 7.0480 \text{ m/s}^2$$ $$a_z \approx -5 + \frac{20}{7.87400787} \approx -5 + 2.53999999 \approx -2.4600 \text{ m/s}^2$$

The x, y, and z components of the net acceleration are:

$$a_x = \boxed{\mathbf{1 + \frac{4}{\sqrt{62}} \approx 1.5080 \text{ m/s}^2}}$$ $$a_y = \boxed{\mathbf{4 + \frac{24}{\sqrt{62}} \approx 7.0480 \text{ m/s}^2}}$$ $$a_z = \boxed{\mathbf{-5 + \frac{20}{\sqrt{62}} \approx -2.4600 \text{ m/s}^2}}$$