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=y^2-2`
C. `x = y^2 + y-6`
D. None of these

Explanation

To solve the problem step by step, we need to find the equation of the path of a particle moving in the x-y plane with given velocities.

### Step 1: Understand the given velocities
We have:
- \( v_x = 8t - 2 \)
- \( v_y = 2 \)

These represent the velocities in the x and y directions, respectively.

### Step 2: Relate velocity to position
The velocity in the x-direction can be expressed as:
\[ v_x = \frac{dx}{dt} \]
Thus, we can write:
\[ \frac{dx}{dt} = 8t - 2 \]

### Step 3: Integrate to find the position in the x-direction
To find the position \( x \), we integrate \( v_x \):
\[ dx = (8t - 2) dt \]
Integrating from \( t = 2 \) to \( t \) and \( x = 14 \) to \( x \):
\[ \int_{14}^{x} dx = \int_{2}^{t} (8t' - 2) dt' \]
Calculating the integral on the right:
\[ x - 14 = \left[ 4t'^2 - 2t' \right]_{2}^{t} \]
Evaluating the limits:
\[ x - 14 = \left( 4t^2 - 2t \right) - \left( 4(2)^2 - 2(2) \right) \]
Calculating \( 4(2)^2 - 2(2) \):
\[ 4(4) - 4 = 16 - 4 = 12 \]
Thus:
\[ x - 14 = 4t^2 - 2t - 12 \]
Rearranging gives:
\[ x = 4t^2 - 2t + 2 \]

### Step 4: Find the position in the y-direction
For the y-direction, we have:
\[ v_y = \frac{dy}{dt} = 2 \]
Integrating:
\[ dy = 2 dt \]
Integrating from \( t = 2 \) to \( t \) and \( y = 4 \) to \( y \):
\[ \int_{4}^{y} dy = \int_{2}^{t} 2 dt' \]
Calculating the integral:
\[ y - 4 = [2t']_{2}^{t} \]
Evaluating the limits:
\[ y - 4 = 2t - 4 \]
Thus:
\[ y = 2t + 4 \]

### Step 5: Express \( t \) in terms of \( y \)
From the equation \( y = 2t + 4 \), we can solve for \( t \):
\[ t = \frac{y - 4}{2} \]

### Step 6: Substitute \( t \) into the equation for \( x \)
Substituting \( t = \frac{y - 4}{2} \) into the equation for \( x \):
\[ x = 4\left(\frac{y - 4}{2}\right)^2 - 2\left(\frac{y - 4}{2}\right) + 2 \]
Calculating:
\[ x = 4\left(\frac{(y - 4)^2}{4}\right) - (y - 4) + 2 \]
This simplifies to:
\[ x = (y - 4)^2 - (y - 4) + 2 \]
Expanding and simplifying:
\[ x = y^2 - 8y + 16 - y + 4 + 2 \]
\[ x = y^2 - 9y + 22 \]

### Final Equation of the Path
Thus, the equation of the path is:
\[ x = y^2 - 9y + 22 \]