A particle is travelling in a circular path of radius 4m. At a certain instant the particle is moving at 20m/s and its acceleration is at an angle of `37^(@)` from the direction to the centre of the circle as seen from the particle
(i) At what rate is the speed of the particle increasing?
(ii) What is the magnitude of the acceleration?

Answer

To solve the problem, we will break it down into two parts as requested:

### Given:
- Radius of the circular path, \( R = 4 \, \text{m} \)
- Speed of the particle, \( V = 20 \, \text{m/s} \)
- Angle of acceleration with respect to the radius (towards the center), \( \theta = 37^\circ \)

### (i) At what rate is the speed of the particle increasing? (Tangential Acceleration)

1. **Calculate the Centripetal Acceleration:**
The centripetal acceleration \( a_n \) is given by the formula:
\[ a_n = \frac{V^2}{R} \]
Substituting the values:
\[ a_n = \frac{20^2}{4} = \frac{400}{4} = 100 \, \text{m/s}^2 \]

2. **Relate Tangential and Centripetal Acceleration using the angle:**
The angle \( \theta \) is given as \( 37^\circ \). The relationship between tangential acceleration \( a_t \) and centripetal acceleration \( a_n \) can be expressed using the tangent of the angle:
\[ \tan(\theta) = \frac{a_t}{a_n} \]
From trigonometric values, we know:
\[ \tan(37^\circ) \approx \frac{3}{4} \]
Thus, we can write:
\[ \frac{3}{4} = \frac{a_t}{100} \]

3. **Solve for Tangential Acceleration \( a_t \):**
Rearranging gives:
\[ a_t = 100 \cdot \frac{3}{4} = 75 \, \text{m/s}^2 \]

### (ii) What is the magnitude of the total acceleration?

1. **Using the Pythagorean theorem:**
The total acceleration \( a \) can be found using the components of the tangential and centripetal accelerations:
\[ a = \sqrt{a_t^2 + a_n^2} \]
Substituting the values:
\[ a = \sqrt{75^2 + 100^2} = \sqrt{5625 + 10000} = \sqrt{15625} \]

2. **Calculate the magnitude:**
\[ a = 125 \, \text{m/s}^2 \]

### Final Answers:
- (i) The rate at which the speed of the particle is increasing (Tangential Acceleration) is \( 75 \, \text{m/s}^2 \).
- (ii) The magnitude of the total acceleration is \( 125 \, \text{m/s}^2 \).