A particle is rotates in a circular path of radius `54m` with varying speed `v=4t^(2)` . Here `v` is in `m//s` and `t` inn second . Find angle between velocity and accelearation at `t=3s` .

Explanation

To find the angle between the velocity and acceleration of a particle rotating in a circular path with varying speed, we can follow these steps:

### Step 1: Determine the speed at t = 3s
Given the speed function:
\[ v = 4t^2 \]
At \( t = 3 \) seconds:
\[ v = 4(3^2) = 4 \times 9 = 36 \, \text{m/s} \]

### Step 2: Calculate the tangential acceleration (\( a_t \))
The tangential acceleration is given by the derivative of the speed with respect to time:
\[ a_t = \frac{dv}{dt} \]
Calculating the derivative:
\[ \frac{dv}{dt} = \frac{d(4t^2)}{dt} = 8t \]
At \( t = 3 \) seconds:
\[ a_t = 8 \times 3 = 24 \, \text{m/s}^2 \]

### Step 3: Calculate the radial acceleration (\( a_r \))
The radial acceleration is given by the formula:
\[ a_r = \frac{v^2}{R} \]
Where \( R = 54 \, \text{m} \) is the radius of the circular path. Substituting the values:
\[ a_r = \frac{(36)^2}{54} = \frac{1296}{54} = 24 \, \text{m/s}^2 \]

### Step 4: Determine the total acceleration (\( a \))
The total acceleration \( a \) can be found using the Pythagorean theorem since \( a_t \) and \( a_r \) are perpendicular:
\[ a = \sqrt{a_t^2 + a_r^2} \]
Substituting the values:
\[ a = \sqrt{(24)^2 + (24)^2} = \sqrt{576 + 576} = \sqrt{1152} = 24\sqrt{2} \, \text{m/s}^2 \]

### Step 5: Find the angle \( \theta \) between velocity and acceleration
Using the relationship between the tangential and radial accelerations:
\[ \tan(\theta) = \frac{a_r}{a_t} \]
Substituting the values:
\[ \tan(\theta) = \frac{24}{24} = 1 \]
Thus,
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Final Answer:
The angle between the velocity and acceleration at \( t = 3 \) seconds is \( 45^\circ \).

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