To solve the problem, we'll first calculate the components of velocity and acceleration in the $x$, $y$, and $z$ directions. Then, we'll determine the velocity and acceleration at $t = 1$. Finally, we'll project these vectors onto the direction vector $\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$.

Step 1: Find the Velocity Components

The velocity vector $\mathbf{v}(t)$ is the derivative of the position vector $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k}$ with respect to time $t$:

Given:

• $x(t) = t^5 + 15$
• $y(t) = 2t^2 - 10$
• $z(t) = 3t + 50$

Compute the derivatives:

• $\frac{dx}{dt} = v_x(t) = 5t^4$
• $\frac{dy}{dt} = v_y(t) = 4t$
• $\frac{dz}{dt} = v_z(t) = 3$

So, the velocity vector is: $\mathbf{v}(t) = 5t^4 \mathbf{i} + 4t \mathbf{j} + 3 \mathbf{k}$

At $t = 1$: $\mathbf{v}(1) = 5(1)^4 \mathbf{i} + 4(1) \mathbf{j} + 3 \mathbf{k} = 5\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}$

Step 2: Find the Acceleration Components

The acceleration vector $\mathbf{a}(t)$ is the derivative of the velocity vector $\mathbf{v}(t)$:

Compute the derivatives:

• $\frac{d^2x}{dt^2} = a_x(t) = 20t^3$
• $\frac{d^2y}{dt^2} = a_y(t) = 4$
• $\frac{d^2z}{dt^2} = a_z(t) = 0$

So, the acceleration vector is: $\mathbf{a}(t) = 20t^3 \mathbf{i} + 4 \mathbf{j} + 0 \mathbf{k}$

At $t = 1$: $\mathbf{a}(1) = 20(1)^3 \mathbf{i} + 4 \mathbf{j} + 0 \mathbf{k} = 20\mathbf{i} + 4\mathbf{j} + 0\mathbf{k}$

Step 3: Direction Vector and Normalization

Given direction vector: $\mathbf{d} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$

Normalize the direction vector: $|\mathbf{d}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$ $\hat{\mathbf{d}} = \frac{1}{3} (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})$

Step 4: Project Velocity and Acceleration onto the Direction Vector

• Velocity projection: $\text{Velocity in direction of } \mathbf{d} = \mathbf{v}(1) \cdot \hat{\mathbf{d}} = (5\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}) \cdot \frac{1}{3}(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})$ $= \frac{1}{3} (5(1) + 4(2) + 3(2)) = \frac{1}{3} (5 + 8 + 6) = \frac{19}{3}$

• Acceleration projection: $\text{Acceleration in direction of } \mathbf{d} = \mathbf{a}(1) \cdot \hat{\mathbf{d}} = (20\mathbf{i} + 4\mathbf{j}) \cdot \frac{1}{3} (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})$ $= \frac{1}{3} (20(1) + 4(2) + 0(2)) = \frac{1}{3} (20 + 8) = \frac{28}{3}$

Final Results:

• Velocity component in the direction $\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ at $t = 1$: $\frac{19}{3}$ units.
• Acceleration component in the direction $\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ at $t = 1$: $\frac{28}{3}$ units.

Would you like more details or have any questions? Here are some related questions to expand your understanding:

1. How do you calculate the magnitude of a vector?
2. What is the physical significance of projecting a vector onto another vector?
3. How does the velocity vector change as the parameter $t$ increases?
4. What is the geometric interpretation of a direction vector?
5. How can you use vectors to describe motion in three-dimensional space?

Tip: When dealing with vector projections, always ensure the direction vector is normalized to maintain accurate component calculations.