If the velocity of a particle moving along x-axis is given as `v=(3t^(2)-2t)` and t=0, x=0 then calculate position of the particle at t=2sec.

Explanation

To find the position of the particle at \( t = 2 \) seconds given the velocity function \( v(t) = 3t^2 - 2t \) and the initial condition \( x(0) = 0 \), we will follow these steps:

### Step 1: Understand the relationship between velocity and position
The velocity \( v \) is the derivative of the position \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} \]
This means that to find the position function \( x(t) \), we need to integrate the velocity function.

### Step 2: Set up the integral
We can express the change in position as:
\[ dx = v \, dt \]
Substituting the given velocity function:
\[ dx = (3t^2 - 2t) \, dt \]

### Step 3: Integrate the velocity function
To find the position function \( x(t) \), we integrate the velocity:
\[ x(t) = \int (3t^2 - 2t) \, dt \]
Calculating the integral:
\[ x(t) = \int (3t^2) \, dt - \int (2t) \, dt \]
\[ x(t) = 3 \cdot \frac{t^3}{3} - 2 \cdot \frac{t^2}{2} + C \]
\[ x(t) = t^3 - t^2 + C \]

### Step 4: Apply the initial condition
We know that at \( t = 0 \), \( x(0) = 0 \):
\[ x(0) = 0^3 - 0^2 + C = 0 \]
This implies:
\[ C = 0 \]
Thus, the position function simplifies to:
\[ x(t) = t^3 - t^2 \]

### Step 5: Calculate the position at \( t = 2 \) seconds
Now we can find the position at \( t = 2 \):
\[ x(2) = 2^3 - 2^2 \]
\[ x(2) = 8 - 4 = 4 \]

### Final Answer
The position of the particle at \( t = 2 \) seconds is:
\[ x(2) = 4 \text{ meters} \]

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