A particle moves in the x-y plane with velocity `v_x = 8t-2 and v_y = 2.` If it passes through the point `x =14 and y = 4` at t = 2 s,` the equation of the path is

Options

A. `x= y^2 -y+2`
B. `x=2y^2+2y-3`
C. `x= 3y^2 +5`
D. can not be found above data

Explanation

To find the equation of the path of the particle moving in the x-y plane with the given velocities, we can follow these steps:

### Step 1: Analyze the y-component of motion
Given that the velocity in the y-direction is constant:
\[ v_y = 2 \]

The displacement in the y-direction can be expressed as:
\[ y = y_0 + v_y \cdot t \]
At \( t = 2 \) seconds, the particle passes through \( y = 4 \):
\[ 4 = y_0 + 2 \cdot 2 \]
\[ 4 = y_0 + 4 \]
From this, we find:
\[ y_0 = 0 \]

### Step 2: Write the equation for y
Now substituting \( y_0 \) back into the equation:
\[ y = 2t \]

### Step 3: Analyze the x-component of motion
The velocity in the x-direction is given by:
\[ v_x = 8t - 2 \]

The relationship between displacement and velocity is:
\[ \frac{dx}{dt} = 8t - 2 \]

### Step 4: Integrate to find x
Integrating both sides with respect to \( t \):
\[ dx = (8t - 2) dt \]
Integrating gives:
\[ x = 4t^2 - 2t + C \]

### Step 5: Determine the constant C
We know that at \( t = 2 \) seconds, \( x = 14 \):
\[ 14 = 4(2^2) - 2(2) + C \]
\[ 14 = 4(4) - 4 + C \]
\[ 14 = 16 - 4 + C \]
\[ 14 = 12 + C \]
Thus:
\[ C = 2 \]

### Step 6: Write the equation for x
Substituting \( C \) back into the equation for \( x \):
\[ x = 4t^2 - 2t + 2 \]

### Step 7: Substitute for t in terms of y
From the equation \( y = 2t \), we can express \( t \) as:
\[ t = \frac{y}{2} \]

### Step 8: Substitute t into the equation for x
Substituting \( t \) into the equation for \( x \):
\[ x = 4\left(\frac{y}{2}\right)^2 - 2\left(\frac{y}{2}\right) + 2 \]
\[ x = 4\left(\frac{y^2}{4}\right) - y + 2 \]
\[ x = y^2 - y + 2 \]

### Final Equation of the Path
Thus, the equation of the path of the particle is:
\[ y^2 - y + 2 = x \]