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, we can use the formula:

\[ R = \frac{V^2}{A_{\perp}} \]

where:
- \( R \) is the radius of curvature,
- \( V \) is the magnitude of the velocity,
- \( A_{\perp} \) is the perpendicular component of the acceleration.

### Step 1: Calculate the magnitude of the velocity vector
The velocity vector is given as:

\[ \vec{V} = 3\hat{i} + 4\hat{j} \]

The magnitude of the velocity \( V \) can be calculated using the formula:

\[ V = \sqrt{V_x^2 + V_y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s} \]

### Step 2: Calculate the magnitude of the acceleration vector
The acceleration vector is given as:

\[ \vec{A} = -25\hat{i} - 25\hat{j} \]

The magnitude of the acceleration \( A \) can be calculated as:

\[ A = \sqrt{A_x^2 + A_y^2} = \sqrt{(-25)^2 + (-25)^2} = \sqrt{625 + 625} = \sqrt{1250} = 25\sqrt{2} \, \text{m/s}^2 \]

### Step 3: Find the angle between the velocity and acceleration vectors
To find the angle \( \theta \) between the velocity and acceleration vectors, we can use the dot product:

\[ \vec{V} \cdot \vec{A} = |\vec{V}| |\vec{A}| \cos(\theta) \]

Calculating the dot product:

\[ \vec{V} \cdot \vec{A} = (3)(-25) + (4)(-25) = -75 - 100 = -175 \]

Now substituting the magnitudes:

\[ -175 = (5)(25\sqrt{2}) \cos(\theta) \]

Solving for \( \cos(\theta) \):

\[ \cos(\theta) = \frac{-175}{125\sqrt{2}} = \frac{-7}{5\sqrt{2}} \]

### Step 4: Calculate the perpendicular component of acceleration
The perpendicular component of acceleration \( A_{\perp} \) can be found using:

\[ A_{\perp} = A \sin(\theta) \]

Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):

\[ \sin^2(\theta) = 1 - \left(\frac{-7}{5\sqrt{2}}\right)^2 = 1 - \frac{49}{50} = \frac{1}{50} \]

Thus,

\[ \sin(\theta) = \frac{1}{\sqrt{50}} = \frac{1}{5\sqrt{2}} \]

Now substituting back to find \( A_{\perp} \):

\[ A_{\perp} = A \sin(\theta) = 25\sqrt{2} \cdot \frac{1}{5\sqrt{2}} = 5 \, \text{m/s}^2 \]

### Step 5: Calculate the radius of curvature
Now we can substitute \( V \) and \( A_{\perp} \) into the radius of curvature formula:

\[ R = \frac{V^2}{A_{\perp}} = \frac{5^2}{5} = \frac{25}{5} = 5 \, \text{m} \]

### Final Answer
The radius of curvature of the trajectory of the particle at that instant is \( \boxed{5 \, \text{m}} \).