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
Velocity is the rate of change of position with respect to time. Mathematically, this is expressed as:
\[ v = \frac{dx}{dt} \]
This implies that:
\[ dx = v \, dt \]

### Step 2: Substitute the velocity function into the equation
We substitute the given velocity function into the equation:
\[ dx = (3t^2 - 2t) \, dt \]

### Step 3: Integrate both sides
To find the position \( x(t) \), we integrate the right side with respect to \( t \):
\[ x(t) = \int (3t^2 - 2t) \, dt \]

### Step 4: Perform the integration
We can integrate each term separately:
\[ x(t) = \int 3t^2 \, dt - \int 2t \, dt \]
Calculating the integrals:
\[ x(t) = 3 \cdot \frac{t^3}{3} - 2 \cdot \frac{t^2}{2} + C \]
This simplifies to:
\[ x(t) = t^3 - t^2 + C \]

### Step 5: Apply the initial condition
We know that when \( t = 0 \), \( x(0) = 0 \). We can use this to find the constant \( C \):
\[ x(0) = 0^3 - 0^2 + C = 0 \implies C = 0 \]
Thus, the position function simplifies to:
\[ x(t) = t^3 - t^2 \]

### Step 6: Calculate the position at \( t = 2 \) seconds
Now, we can find the position at \( t = 2 \):
\[ x(2) = 2^3 - 2^2 = 8 - 4 = 4 \]

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

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