The relation between time and displacement x is `t =alphax^(2) + betax`, where `alpha` and `beta` are constants. (Take `alpha =2 m^(-2) s, beta = 1m^(-1)s)`

Options

A. The relation between velocity and acceleration is `-3//5 nualpha^(2)`
B. Atx= 2 the velocity is 3/5 m `s^(-1)`
C. The relation between velocity and acceleration is` -2 alphanu^(3)` .
D. At x=2 the velocity is `(1)/(9) ms^(-1)`

Explanation

To solve the problem, we need to find the relationship between velocity and acceleration given the equation for time as a function of displacement:

\[ t = \alpha x^2 + \beta x \]

where \( \alpha = 2 \, \text{m}^{-2} \text{s} \) and \( \beta = 1 \, \text{m}^{-1} \text{s} \).

### Step-by-Step Solution:

1. **Differentiate Time with Respect to Displacement**:
We start by differentiating the equation for time \( t \) with respect to displacement \( x \):
\[ \frac{dt}{dx} = \frac{d}{dx}(\alpha x^2 + \beta x) = 2\alpha x + \beta \]

2. **Find Velocity**:
The velocity \( v \) is defined as the rate of change of displacement with respect to time, which is the reciprocal of the rate of change of time with respect to displacement:
\[ v = \frac{dx}{dt} = \frac{1}{\frac{dt}{dx}} = \frac{1}{2\alpha x + \beta} \]

3. **Substitute Values for \( \alpha \) and \( \beta \)**:
Now we substitute the values of \( \alpha \) and \( \beta \):
\[ v = \frac{1}{2(2)x + 1} = \frac{1}{4x + 1} \]

4. **Calculate Velocity at \( x = 2 \)**:
We need to find the velocity when \( x = 2 \):
\[ v = \frac{1}{4(2) + 1} = \frac{1}{8 + 1} = \frac{1}{9} \, \text{m/s} \]

5. **Find Acceleration**:
To find acceleration \( a \), we differentiate the velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \]

6. **Differentiate Velocity with Respect to Displacement**:
We first differentiate \( v = \frac{1}{4x + 1} \):
\[ \frac{dv}{dx} = -\frac{4}{(4x + 1)^2} \]

7. **Substitute \( \frac{dv}{dx} \) into the Acceleration Equation**:
Now substituting back into the acceleration equation:
\[ a = \left(-\frac{4}{(4x + 1)^2}\right) \cdot v \]
Substituting \( v = \frac{1}{4x + 1} \):
\[ a = -\frac{4}{(4x + 1)^2} \cdot \frac{1}{4x + 1} = -\frac{4}{(4x + 1)^3} \]

8. **Relate Acceleration to Velocity**:
We can express acceleration in terms of velocity:
\[ a = -\frac{4v}{(4x + 1)^2} \]
Since \( v = \frac{1}{4x + 1} \), we can substitute \( v \) back to express acceleration in terms of \( v \):
\[ a = -4v^3 \]

### Final Result:
The relationship between acceleration \( a \) and velocity \( v \) is given by:
\[ a = -2\alpha v^3 \]
where \( \alpha = 2 \, \text{m}^{-2} \text{s} \).