It is given that `t= px^(2) + qx`, where x is displacement and t is time. The acceleration of particle at origin is

Options

A. `-(2p)/(q^(3))`
B. `-(2q)/(p^(3))`
C. `(2p)/(q^(3))`
D. `(2q)/(p^(3))`

Explanation

To find the acceleration of a particle at the origin given the equation \( t = px^2 + qx \), we will follow these steps:

### Step 1: Differentiate the given equation with respect to time
We start with the equation:
\[ t = px^2 + qx \]
Differentiating both sides with respect to \( t \):
\[ \frac{dt}{dt} = \frac{d(px^2 + qx)}{dt} \]
This simplifies to:
\[ 1 = 2px \frac{dx}{dt} + q \frac{dx}{dt} \]

### Step 2: Factor out \(\frac{dx}{dt}\)
We can factor out \(\frac{dx}{dt}\) from the right-hand side:
\[ 1 = \left(2px + q\right) \frac{dx}{dt} \]

### Step 3: Differentiate again to find acceleration
Now, we differentiate the equation \( 1 = (2px + q) \frac{dx}{dt} \) again with respect to \( t \):
\[ 0 = \frac{d}{dt}\left((2px + q) \frac{dx}{dt}\right) \]
Using the product rule:
\[ 0 = \left(2p \frac{dx}{dt} + 2px \frac{d^2x}{dt^2}\right) + q \frac{d^2x}{dt^2} \]
This simplifies to:
\[ 0 = 2p \frac{dx}{dt} + (2px + q) \frac{d^2x}{dt^2} \]

### Step 4: Substitute \( x = 0 \) to find acceleration at the origin
At the origin, set \( x = 0 \):
\[ 0 = 2p \frac{dx}{dt} + q \frac{d^2x}{dt^2} \]
This implies:
\[ 0 = 2p \cdot v_0 + q \cdot a_0 \]
where \( v_0 = \frac{dx}{dt} \) and \( a_0 = \frac{d^2x}{dt^2} \).

### Step 5: Solve for \( v_0 \)
From the first differentiation step, substituting \( x = 0 \):
\[ 1 = (2p \cdot 0 + q) v_0 \implies 1 = q v_0 \implies v_0 = \frac{1}{q} \]

### Step 6: Substitute \( v_0 \) back to find \( a_0 \)
Substituting \( v_0 \) back into the equation for acceleration:
\[ 0 = 2p \left(\frac{1}{q}\right) + q a_0 \]
This simplifies to:
\[ 0 = \frac{2p}{q} + q a_0 \]
Rearranging gives:
\[ q a_0 = -\frac{2p}{q} \implies a_0 = -\frac{2p}{q^2} \]

### Final Result
Thus, the acceleration of the particle at the origin is:
\[ a_0 = -\frac{2p}{q^2} \]