The relation between time `t` and distance `x` is `t = ax^(2)+ bx` where `a and `b` are constants. The acceleration is

Options

A. ` 2 bv^(3)`
B. ` - 2 abv^(2)`
C. ` 2 av^(2)`
D. ` -2 av^(3)`

Explanation

To find the acceleration from the given relation between time \( t \) and distance \( x \) expressed as \( t = ax^2 + bx \), we will follow these steps:

### Step 1: Differentiate the equation with respect to time \( t \)
We start with the equation:
\[ t = ax^2 + bx \]
To find the velocity, we differentiate both sides with respect to \( t \):
\[ \frac{dt}{dt} = \frac{d}{dt}(ax^2 + bx) \]
This simplifies to:
\[ 1 = \frac{d}{dt}(ax^2) + \frac{d}{dt}(bx) \]

### Step 2: Apply the chain rule
Using the chain rule for differentiation:
\[ \frac{d}{dt}(ax^2) = 2ax \frac{dx}{dt} \quad \text{and} \quad \frac{d}{dt}(bx) = b \frac{dx}{dt} \]
Thus, we have:
\[ 1 = 2ax \frac{dx}{dt} + b \frac{dx}{dt} \]

### Step 3: Factor out the velocity
Let \( v = \frac{dx}{dt} \) (the velocity). We can rewrite the equation as:
\[ 1 = (2ax + b)v \]
Now, we can express \( v \):
\[ v = \frac{1}{2ax + b} \]

### Step 4: Differentiate \( v \) with respect to \( t \) to find acceleration
The acceleration \( a \) is defined as:
\[ a = \frac{dv}{dt} \]
Using the chain rule again, we have:
\[ \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \]
Now we need to find \( \frac{dv}{dx} \):
\[ v = (2ax + b)^{-1} \]
Differentiating \( v \) with respect to \( x \):
\[ \frac{dv}{dx} = -\frac{1}{(2ax + b)^2} \cdot \frac{d}{dx}(2ax + b) = -\frac{1}{(2ax + b)^2} \cdot 2a \]
Thus,
\[ \frac{dv}{dx} = -\frac{2a}{(2ax + b)^2} \]

### Step 5: Substitute back to find acceleration
Now substituting \( \frac{dv}{dx} \) into the acceleration formula:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot v = -\frac{2a}{(2ax + b)^2} \cdot \frac{1}{(2ax + b)} \]
This simplifies to:
\[ a = -\frac{2a}{(2ax + b)^3} \]

### Final Result
Thus, the acceleration is:
\[ a = -\frac{2a}{(2ax + b)^3} \]