The reation between the time t and position x for a particle moving on x-axis is given by `t=px^(2)+qx`, where p and q are constants. The relation between velocity v and acceleration a is as

Options

A. `a prop v^(3)`
B. `a prop v^(2)`
C. `a prop v^(4)`
D. `a prop v`

Answer

To find the relationship between velocity \( v \) and acceleration \( a \) for a particle moving along the x-axis with the given time-position relation \( t = px^2 + qx \), we will follow these steps:

### Step 1: Differentiate the time-position relation
Given:
\[ t = px^2 + qx \]

We differentiate both sides with respect to time \( t \):
\[ \frac{dt}{dt} = \frac{d}{dt}(px^2 + qx) \]

This simplifies to:
\[ 1 = \frac{d}{dt}(px^2) + \frac{d}{dt}(qx) \]

### Step 2: Apply the chain rule
Using the chain rule, we have:
\[ \frac{d}{dt}(px^2) = p \cdot 2x \cdot \frac{dx}{dt} = 2pxv \]
\[ \frac{d}{dt}(qx) = q \cdot \frac{dx}{dt} = qv \]

Thus, the equation becomes:
\[ 1 = 2pxv + qv \]

### Step 3: Solve for \( x \)
Rearranging the equation gives:
\[ 2pxv + qv = 1 \]
Factoring out \( v \):
\[ v(2px + q) = 1 \]

Now, we can solve for \( x \):
\[ 2px + q = \frac{1}{v} \]
\[ 2px = \frac{1}{v} - q \]
\[ x = \frac{1 - qv}{2pv} \]

### Step 4: Differentiate the velocity to find acceleration
Now we need to find acceleration \( a \), which is the derivative of velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} \]

We differentiate the expression for \( v \) with respect to \( t \):
Using \( v(2px + q) = 1 \), we differentiate both sides:
\[ \frac{d}{dt}(v(2px + q)) = 0 \]

Using the product rule:
\[ \frac{dv}{dt}(2px + q) + v\frac{d}{dt}(2px + q) = 0 \]

### Step 5: Differentiate \( 2px + q \)
Now we differentiate \( 2px + q \):
\[ \frac{d}{dt}(2px + q) = 2p\left(\frac{dx}{dt}\right) = 2pv \]

Substituting back into the equation:
\[ \frac{dv}{dt}(2px + q) + v(2pv) = 0 \]

### Step 6: Substitute \( x \) back into the equation
Substituting \( x = \frac{1 - qv}{2pv} \) into the equation:
\[ \frac{dv}{dt}(2p\left(\frac{1 - qv}{2pv}\right) + q) + 2pv^2 = 0 \]

### Step 7: Simplify the equation
After simplification, we will find a relationship between \( a \) and \( v \):
\[ a = -2pvq \]

### Conclusion
Thus, the relationship between acceleration \( a \) and velocity \( v \) is:
\[ a \propto -v \]