Two particles move along an `x` axis. The position of particle 1 is given by `x=6.00t^(2) + 3.00 t + 2.00` ( in meters and seconds ) , the acceleration of particle 2 is given by `a= -8.00t` ( in meters per second squared and seconds ) and , at `t=0`, its velocity is . 15 m/s . When the velocities of the particles match, what is their velocity ?

Answer

To solve the problem step by step, we will first find the velocities of both particles and then determine the time at which their velocities match. Finally, we will calculate the common velocity at that time.

### Step 1: Find the velocity of Particle 1
The position of Particle 1 is given by the equation:
\[ x_1 = 6.00t^2 + 3.00t + 2.00 \]

To find the velocity \( v_1 \), we differentiate the position with respect to time \( t \):
\[ v_1 = \frac{dx_1}{dt} = \frac{d}{dt}(6.00t^2 + 3.00t + 2.00) \]

Calculating the derivative:
\[ v_1 = 12.00t + 3.00 \]

### Step 2: Find the velocity of Particle 2
The acceleration of Particle 2 is given by:
\[ a_2 = -8.00t \]

We know that acceleration is the derivative of velocity with respect to time:
\[ a_2 = \frac{dv_2}{dt} \]

Thus, we can express the change in velocity as:
\[ dv_2 = a_2 dt = -8.00t dt \]

Integrating both sides, we have:
\[ \int dv_2 = \int -8.00t dt \]
\[ v_2 - v_{2,0} = -4.00t^2 \]

Given that the initial velocity \( v_{2,0} \) at \( t = 0 \) is 15 m/s:
\[ v_2 - 15 = -4.00t^2 \]
\[ v_2 = -4.00t^2 + 15 \]

### Step 3: Set the velocities equal to find the time
To find when the velocities of both particles are equal, we set \( v_1 = v_2 \):
\[ 12.00t + 3.00 = -4.00t^2 + 15 \]

Rearranging the equation:
\[ 4.00t^2 + 12.00t - 12.00 = 0 \]

### Step 4: Solve the quadratic equation
We can simplify the equation by dividing by 4:
\[ t^2 + 3t - 3 = 0 \]

Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 3, c = -3 \):
\[ t = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \]
\[ t = \frac{-3 \pm \sqrt{9 + 12}}{2} \]
\[ t = \frac{-3 \pm \sqrt{21}}{2} \]

Since time cannot be negative, we take the positive root:
\[ t = \frac{-3 + \sqrt{21}}{2} \]

### Step 5: Calculate the common velocity
Now we substitute \( t \) back into either \( v_1 \) or \( v_2 \) to find the common velocity. Using \( v_1 \):
\[ v_1 = 12.00\left(\frac{-3 + \sqrt{21}}{2}\right) + 3.00 \]

Calculating this:
\[ v_1 = 6.00(-3 + \sqrt{21}) + 3.00 \]
\[ v_1 = -18 + 6\sqrt{21} + 3 \]
\[ v_1 = 6\sqrt{21} - 15 \]

Calculating \( 6\sqrt{21} \) numerically:
\[ \sqrt{21} \approx 4.58 \]
\[ 6\sqrt{21} \approx 27.48 \]
Thus:
\[ v_1 \approx 27.48 - 15 = 12.48 \text{ m/s} \]

### Final Answer
The common velocity when the velocities of the two particles match is approximately:
\[ \boxed{12.48 \text{ m/s}} \]