The velocity of a particle moving on the x-axis is given by `v=x^(2)+x`, where x is in m and v in m/s. What is its position (in m) when its acceleration is `30m//s^(2)`.

Answer

To solve the problem, we need to find the position \( x \) when the acceleration \( a \) of the particle is \( 30 \, \text{m/s}^2 \). The velocity \( v \) is given as a function of position \( x \) by the equation:

\[ v = x^2 + x \]

### Step 1: Express acceleration in terms of position
We know that acceleration can be expressed in terms of velocity and position as follows:

\[ a = v \frac{dv}{dx} \]

### Step 2: Differentiate the velocity function
First, we need to differentiate the velocity function \( v = x^2 + x \) with respect to \( x \):

\[ \frac{dv}{dx} = \frac{d}{dx}(x^2 + x) = 2x + 1 \]

### Step 3: Substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration formula
Now we can substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration formula:

\[ a = (x^2 + x)(2x + 1) \]

### Step 4: Set the acceleration equal to \( 30 \, \text{m/s}^2 \)
We set the expression for acceleration equal to \( 30 \):

\[ (x^2 + x)(2x + 1) = 30 \]

### Step 5: Expand and rearrange the equation
Expanding the left-hand side:

\[ 2x^3 + x^2 + 2x^2 + x = 30 \]

This simplifies to:

\[ 2x^3 + 3x^2 + x - 30 = 0 \]

### Step 6: Solve the cubic equation
Now we need to solve the cubic equation \( 2x^3 + 3x^2 + x - 30 = 0 \). We can use trial and error to find rational roots.

1. Testing \( x = 1 \):
\[ 2(1)^3 + 3(1)^2 + (1) - 30 = 2 + 3 + 1 - 30 = -24 \quad (\text{not a root}) \]

2. Testing \( x = 2 \):
\[ 2(2)^3 + 3(2)^2 + (2) - 30 = 16 + 12 + 2 - 30 = 0 \quad (\text{is a root}) \]

### Step 7: Conclusion
Thus, we found that \( x = 2 \) meters is the position when the acceleration is \( 30 \, \text{m/s}^2 \).

### Final Answer
The position \( x \) when the acceleration is \( 30 \, \text{m/s}^2 \) is:

\[ \boxed{2 \, \text{m}} \]