A particle is moving in such a way that at one instant velocity vector of the particle is `3hati+4hatj m//s` and acceleration vector is `-25hati - 25hatj m//s`? The radius of curvature of the trajectory of the particle at that instant is

Options

A. 2 m
B. 3 m
C. 5 m
D. 10 m

Answer

To find the radius of curvature of the trajectory of the particle given its velocity and acceleration vectors, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Vectors**:
- Velocity vector \( \mathbf{v} = 3 \hat{i} + 4 \hat{j} \) m/s
- Acceleration vector \( \mathbf{a} = -25 \hat{i} - 25 \hat{j} \) m/s²

2. **Calculate the Magnitude of Velocity**:
\[ |\mathbf{v}| = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ m/s} \]

3. **Determine the Angle of the Velocity Vector**:
- The angle \( \theta \) with respect to the horizontal can be calculated using:
\[ \tan(\theta) = \frac{4}{3} \]
- Therefore, \( \theta = \tan^{-1}\left(\frac{4}{3}\right) \).

4. **Find the Components of Acceleration**:
- The acceleration vector can be broken down into components:
- \( a_x = -25 \) m/s²
- \( a_y = -25 \) m/s²

5. **Calculate the Perpendicular Component of Acceleration**:
- The angle of the velocity vector is \( 53^\circ \) (from the previous calculation).
- The angle of the acceleration vector is \( 53^\circ + 90^\circ = 143^\circ \).
- The perpendicular component of the acceleration can be calculated using:
\[ a_{\perpendicular} = |\mathbf{a}| \cdot \sin(37^\circ) \]
- Where \( 37^\circ \) is the complementary angle to \( 53^\circ \):
\[ |\mathbf{a}| = \sqrt{(-25)^2 + (-25)^2} = \sqrt{625 + 625} = \sqrt{1250} = 25\sqrt{2} \text{ m/s}^2 \]
- The sine of \( 37^\circ \) is \( \frac{3}{5} \):
\[ a_{\perpendicular} = 25 \cdot \frac{3}{5} = 15 \text{ m/s}^2 \]

6. **Calculate the Radius of Curvature**:
- The radius of curvature \( R \) can be calculated using the formula:
\[ R = \frac{v^2}{a_{\perpendicular}} \]
- Substituting the values:
\[ R = \frac{(5)^2}{15} = \frac{25}{15} = \frac{5}{3} \text{ m} \]

### Final Answer:
The radius of curvature of the trajectory of the particle at that instant is \( \frac{5}{3} \) m.