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To solve this problem, we need to determine the magnitudes of the velocity and acceleration of point \( P \) based on the provided kinematic relationships.

### Given Data:
- The vertical position \( y = 150 \, \text{mm} \)
- The velocity in the vertical direction \( \frac{dy}{dt} = 300 \, \text{mm/s} \)
- The vertical acceleration \( \frac{d^2y}{dt^2} = 0 \, \text{mm/s}^2 \) (i.e., no vertical acceleration)

The rod moves through a curved slot, and point \( P \) follows this path. The curved slot has a radius of \( 300 \, \text{mm} \).

### Solution Process:

#### 1. Velocity of Point P:
The velocity of point \( P \) consists of two components:
- Vertical component \( v_y = \frac{dy}{dt} = 300 \, \text{mm/s} \).
- Horizontal component \( v_x \) due to the movement along the curved path.

The curved path gives a relationship between the horizontal displacement and the vertical displacement. Using the radius \( R = 300 \, \text{mm} \), we can express the relationship between the components using the geometric constraints of the slot.

For small movements along the curved slot:
\[
 R^2 = x^2 + y^2
\]
Taking the time derivative of both sides with respect to time:
\[
 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
\]
Simplifying:
\[
 x \frac{dx}{dt} = -y \frac{dy}{dt}
\]
Thus, the horizontal velocity \( v_x = \frac{dx}{dt} \) is given by:
\[
 v_x = -\frac{y}{x} \cdot v_y
\]
Now, we need to find \( x \). From the geometry:
\[
 x = \sqrt{R^2 - y^2} = \sqrt{300^2 - 150^2} = \sqrt{90000 - 22500} = \sqrt{67500} \, \text{mm}
\]
\[
 x \approx 259.81 \, \text{mm}
\]
Now, calculate \( v_x \):
\[
 v_x = -\frac{150}{259.81} \times 300 \approx -173.14 \, \text{mm/s}
\]

#### 2. Magnitude of the Total Velocity:
The total velocity is the magnitude of the velocity vector, combining \( v_x \) and \( v_y \):
\[
 v = \sqrt{v_x^2 + v_y^2} = \sqrt{(-173.14)^2 + 300^2}
\]
\[
 v \approx \sqrt{29979.71 + 90000} = \sqrt{119979.71} \approx 346.38 \, \text{mm/s}
\]

#### 3. Acceleration of Point P:
Since \( \frac{d^2y}{dt^2} = 0 \), the vertical acceleration is zero. However, there may be a radial or centripetal acceleration due to the curved path.

The radial acceleration \( a_r \) is given by:
\[
 a_r = \frac{v^2}{R}
\]
Substituting:
\[
 a_r = \frac{346.38^2}{300} = \frac{119979.71}{300} \approx 399.93 \, \text{mm/s}^2
\]

#### Final Answer:
- **Velocity magnitude** of point \( P \): \( 346.38 \, \text{mm/s} \)
- **Acceleration magnitude** of point \( P \): \( 399.93 \, \text{mm/s}^2 \)

Do you need more details or clarifications?

### Related Questions:
1. How would the result change if the radius of the curve was different?
2. What happens to the velocity and acceleration if the vertical acceleration \( \frac{d^2y}{dt^2} \neq 0 \)?
3. Can we solve this problem for a different curved path, such as a parabolic slot?
4. What forces are acting on point \( P \) as it moves along the curve?
5. How do the velocity and acceleration change if the initial vertical velocity \( \frac{dy}{dt} \) was zero?

### Tip:
Remember that in curved motion, the acceleration always has a radial component related to the velocity and the curvature of the path, even if there is no change in vertical velocity.

If y = 150 mm, dy/dt = 300 mm/s, and d²y/dt² = 0, what are the magnitudes of the velocity and acceleration of point P?

Learn how to calculate the magnitudes of velocity and acceleration for a point moving along a curved slot. This problem involves the kinematics of point P, with detailed steps including key formulas for velocity, radial acceleration, and geometric constraints of circular motion.

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