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We are given the position of a particle in the xy-plane as:

\[
r(t) = (t - 4)\hat{i} + (t^2 + 1)\hat{j}
\]

Our goal is to find:
1. The equation for the path of the particle in terms of \(y\) and \(x\).
2. The velocity vector at \(t = 5\).
3. The acceleration vector at \(t = 5\).

### Step 1: Equation for the path of the particle
We are given:
- \(x(t) = t - 4\)
- \(y(t) = t^2 + 1\)

To eliminate \(t\), solve for \(t\) in terms of \(x\):
\[
x = t - 4 \implies t = x + 4
\]

Substitute \(t = x + 4\) into the equation for \(y\):
\[
y = (x + 4)^2 + 1
\]
Simplify:
\[
y = x^2 + 8x + 16 + 1 \implies y = x^2 + 8x + 17
\]

Thus, the equation for the path is:
\[
y = x^2 + 8x + 17
\]

### Step 2: Velocity vector at \(t = 5\)
The velocity vector is the derivative of the position vector:
\[
v(t) = \frac{d}{dt}[(t - 4)\hat{i} + (t^2 + 1)\hat{j}]
\]
Differentiate each component:
\[
v(t) = \left(1\hat{i} + 2t\hat{j}\right)
\]

At \(t = 5\):
\[
v(5) = 1\hat{i} + 2(5)\hat{j} = 1\hat{i} + 10\hat{j}
\]

Thus, the velocity vector at \(t = 5\) is:
\[
v = 1\hat{i} + 10\hat{j}
\]

### Step 3: Acceleration vector at \(t = 5\)
The acceleration vector is the derivative of the velocity vector:
\[
a(t) = \frac{d}{dt}[1\hat{i} + 2t\hat{j}] = 0\hat{i} + 2\hat{j}
\]

At \(t = 5\):
\[
a(5) = 0\hat{i} + 2\hat{j}
\]

Thus, the acceleration vector at \(t = 5\) is:
\[
a = 0\hat{i} + 2\hat{j}
\]

### Final Answers:
- The equation for the path is: \(y = x^2 + 8x + 17\).
- The velocity vector at \(t = 5\) is: \(1\hat{i} + 10\hat{j}\).
- The acceleration vector at \(t = 5\) is: \(0\hat{i} + 2\hat{j}\).

---

Would you like further details on any step?

Here are 5 related questions:
1. How do we derive the velocity and acceleration vectors from the position vector in 3D?
2. What does the equation of the path tell us about the shape of the particle's trajectory?
3. How would the velocity and acceleration vectors change for different values of \(t\)?
4. How do velocity and acceleration vectors relate to the speed and direction of the particle?
5. Can we determine the particle's speed at \(t = 5\)?

**Tip:** The velocity vector provides both the speed and direction of motion. You can calculate speed by finding the magnitude of the velocity vector.

The position of a particle in the xy-plane at time t is r(t) = (t - 4)i + (t^2 + 1)j. Find an equation in x and y whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at t = 5.

Explore how to derive the path equation y = x^2 + 8x + 17 for a particle moving in the xy-plane with position r(t) = (t - 4)i + (t^2 + 1)j. Learn how to calculate the velocity vector v = 1i + 10j and the acceleration vector a = 0i + 2j at t = 5. This problem is ideal for Grades 11-12 students studying vector calculus and kinematics.

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