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 solve the problem, we need to find the relationship between velocity \( v \) and acceleration \( a \) for a particle whose position \( x \) is related to time \( t \) by the equation:

\[ t = px^2 + qx \]

where \( p \) and \( q \) are constants.

### Step-by-Step Solution

**Step 1: Differentiate the given equation with respect to time \( t \)**

Starting with the equation:

\[ t = px^2 + qx \]

Differentiating both sides with respect to \( t \):

\[ \frac{dt}{dt} = \frac{d}{dt}(px^2 + qx) \]

This simplifies to:

\[ 1 = 2px \frac{dx}{dt} + q \frac{dx}{dt} \]

**Step 2: Express \( \frac{dx}{dt} \) in terms of \( v \)**

Since \( \frac{dx}{dt} = v \) (velocity), we can rewrite the equation as:

\[ 1 = (2px + q)v \]

**Step 3: Solve for \( v \)**

Rearranging gives:

\[ v = \frac{1}{2px + q} \]

**Step 4: Differentiate \( v \) with respect to \( x \)**

To find acceleration \( a \), we need to differentiate \( v \) with respect to \( t \):

Using the chain rule:

\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \]

**Step 5: Differentiate \( v \) with respect to \( x \)**

Now, we differentiate \( v = \frac{1}{2px + q} \) with respect to \( x \):

Using the quotient rule:

\[ \frac{dv}{dx} = -\frac{(2p)(1)}{(2px + q)^2} = -\frac{2p}{(2px + q)^2} \]

**Step 6: Substitute \( \frac{dv}{dx} \) back into the acceleration equation**

Now substituting back into the acceleration formula:

\[ a = \frac{dv}{dx} \cdot v = -\frac{2p}{(2px + q)^2} \cdot \frac{1}{2px + q} \]

This simplifies to:

\[ a = -\frac{2p}{(2px + q)^3} \]

**Step 7: Relate \( a \) and \( v \)**

From our expression for \( v \):

\[ v = \frac{1}{2px + q} \]

We can express \( (2px + q) \) in terms of \( v \):

\[ 2px + q = \frac{1}{v} \]

Substituting this back into the expression for \( a \):

\[ a = -2p v^3 \]

Thus, we can conclude that:

\[ a \propto -v^3 \]

### Final Result

The relationship between acceleration \( a \) and velocity \( v \) is:

\[ a \propto v^3 \]