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

Options

A. (a)`2av^(2) `
B. (b)` -2av^(3)`
C. (c)` abv^(2)`
D. (d)` -2abv^(2)`

Explanation

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

### Step 1: Differentiate the equation with respect to \( x \)
Given:
\[ t = ax^2 + bx \]
We differentiate both sides with respect to \( x \):
\[ \frac{dt}{dx} = 2ax + b \]

### Step 2: Find the expression for velocity \( v \)
We know that velocity \( v \) is defined as:
\[ v = \frac{dx}{dt} \]
Taking the reciprocal of the derivative we found in Step 1:
\[ v = \frac{dx}{dt} = \frac{1}{\frac{dt}{dx}} = \frac{1}{2ax + b} \]

### Step 3: Differentiate velocity with respect to time to find acceleration
Acceleration \( a \) is defined as:
\[ a = \frac{dv}{dt} \]
Using the chain rule, we can express this as:
\[ a = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
We already have \( \frac{dx}{dt} = v \). Now we need to find \( \frac{dv}{dx} \).

### Step 4: Differentiate \( v \) with respect to \( x \)
From Step 2:
\[ v = \frac{1}{2ax + b} \]
Now differentiate \( v \) with respect to \( x \):
\[ \frac{dv}{dx} = -\frac{2a}{(2ax + b)^2} \]

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

### Final Expression for Acceleration
Thus, the acceleration is given by:
\[ a = -2A v^3 \]