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 =2s the equation of the path is

Options

A. 1. `x=y^3-y^2+2`
B. 2. `x=y^2 - y+2`
C. 3. `x=y^2 - 3y+2`
D. 4. `x=y^3 - 2y^2 +2`

Answer

To solve the problem, we need to find the equation of the path of a particle moving in the x-y plane given its velocity components. Let's break down the solution step by step.

### Step 1: Write down the given equations
We have the velocity components:
- \( v_x = 8t - 2 \)
- \( v_y = 2 \)

Also, the particle passes through the point \( (x, y) = (14, 4) \) at \( t = 2 \) seconds.

### Step 2: Express the velocities in differential form
The velocities can be expressed in differential form:
- \( v_x = \frac{dx}{dt} = 8t - 2 \)
- \( v_y = \frac{dy}{dt} = 2 \)

### Step 3: Integrate the x-component of velocity
We will integrate \( v_x \) to find \( x \):
\[ dx = (8t - 2) dt \]
Integrating both sides:
\[ \int dx = \int (8t - 2) dt \]
This gives:
\[ x = 4t^2 - 2t + C_x \]
where \( C_x \) is the constant of integration.

### Step 4: Determine the constant of integration for x
We know that at \( t = 2 \), \( x = 14 \):
\[ 14 = 4(2^2) - 2(2) + C_x \]
Calculating:
\[ 14 = 4(4) - 4 + C_x \]
\[ 14 = 16 - 4 + C_x \]
\[ 14 = 12 + C_x \implies C_x = 2 \]
Thus, the equation for \( x \) becomes:
\[ x = 4t^2 - 2t + 2 \quad \text{(Equation 1)} \]

### Step 5: Integrate the y-component of velocity
Now, we integrate \( v_y \):
\[ dy = 2 dt \]
Integrating gives:
\[ \int dy = \int 2 dt \]
This results in:
\[ y = 2t + C_y \]
where \( C_y \) is another constant of integration.

### Step 6: Determine the constant of integration for y
Using the condition at \( t = 2 \), \( y = 4 \):
\[ 4 = 2(2) + C_y \]
Calculating:
\[ 4 = 4 + C_y \implies C_y = 0 \]
Thus, the equation for \( y \) becomes:
\[ y = 2t \quad \text{(Equation 2)} \]

### Step 7: Eliminate t to find the equation of the path
From Equation 2, we can express \( t \) in terms of \( y \):
\[ t = \frac{y}{2} \]
Now substitute this into Equation 1:
\[ x = 4\left(\frac{y}{2}\right)^2 - 2\left(\frac{y}{2}\right) + 2 \]
Simplifying:
\[ x = 4\left(\frac{y^2}{4}\right) - y + 2 \]
\[ x = y^2 - y + 2 \]

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