A particle's velocity `v` at time`t` is given by `v=2e^(2t)cos((pit)/3)`.the least value of `t` for which the acceleration becomes zero is

Options

A. 0
B. `3/2`
C. `3/pi tan^(-1) (6/pi)`
D. `3/pi cot^(-1) (6/pi)`

Answer

To find the least value of \( t \) for which the acceleration of the particle becomes zero, we follow these steps:

### Step 1: Write down the given velocity function
The velocity \( v \) of the particle at time \( t \) is given by:
\[ v = 2e^{2t} \cos\left(\frac{\pi t}{3}\right) \]

### Step 2: Differentiate the velocity to find acceleration
Acceleration \( a \) is defined as the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
To differentiate \( v \), we will use the product rule:
\[ \frac{dv}{dt} = \frac{d}{dt}(2e^{2t}) \cdot \cos\left(\frac{\pi t}{3}\right) + 2e^{2t} \cdot \frac{d}{dt}\left(\cos\left(\frac{\pi t}{3}\right)\right) \]

### Step 3: Compute the derivatives
1. Differentiate \( 2e^{2t} \):
\[ \frac{d}{dt}(2e^{2t}) = 2e^{2t} \cdot 2 = 4e^{2t} \]

2. Differentiate \( \cos\left(\frac{\pi t}{3}\right) \):
\[ \frac{d}{dt}\left(\cos\left(\frac{\pi t}{3}\right)\right) = -\sin\left(\frac{\pi t}{3}\right) \cdot \frac{\pi}{3} \]

### Step 4: Substitute the derivatives back into the acceleration formula
Now substituting these derivatives back:
\[ a = 4e^{2t} \cos\left(\frac{\pi t}{3}\right) - 2e^{2t} \cdot \sin\left(\frac{\pi t}{3}\right) \cdot \frac{\pi}{3} \]
Factoring out \( 2e^{2t} \):
\[ a = 2e^{2t} \left(2 \cos\left(\frac{\pi t}{3}\right) - \frac{\pi}{3} \sin\left(\frac{\pi t}{3}\right)\right) \]

### Step 5: Set acceleration to zero
To find when the acceleration is zero, we set the expression inside the parentheses to zero:
\[ 2 \cos\left(\frac{\pi t}{3}\right) - \frac{\pi}{3} \sin\left(\frac{\pi t}{3}\right) = 0 \]
Rearranging gives:
\[ \frac{2 \cos\left(\frac{\pi t}{3}\right)}{\sin\left(\frac{\pi t}{3}\right)} = \frac{\pi}{3} \]
This simplifies to:
\[ \tan\left(\frac{\pi t}{3}\right) = \frac{6}{\pi} \]

### Step 6: Solve for \( t \)
Taking the arctangent of both sides:
\[ \frac{\pi t}{3} = \tan^{-1}\left(\frac{6}{\pi}\right) \]
Thus,
\[ t = \frac{3}{\pi} \tan^{-1}\left(\frac{6}{\pi}\right) \]

### Final Answer
The least value of \( t \) for which the acceleration becomes zero is:
\[ t = \frac{3}{\pi} \tan^{-1}\left(\frac{6}{\pi}\right) \]