A particle move a distance `x` in time `t` according to equation `x = (t + 5)^-1`. The acceleration of particle is alphaortional to.

Options

A. `("velocity")^(3//2)`
B. `("distance")^2`
C. `("distance")^-2`
D. `("velocity")^(2//3)`

Explanation

To solve the problem step by step, we will follow the process of finding the velocity and acceleration from the given displacement equation.

### Step 1: Write down the displacement equation
The displacement of the particle is given by:
\[ x = (t + 5)^{-1} \]

### Step 2: Differentiate the displacement to find velocity
To find the velocity \( v \), we differentiate \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} \]
Using the chain rule, we differentiate:
\[ v = \frac{d}{dt} \left( (t + 5)^{-1} \right) \]
Applying the power rule:
\[ v = -1 \cdot (t + 5)^{-2} \cdot \frac{d}{dt}(t + 5) \]
Since \( \frac{d}{dt}(t + 5) = 1 \):
\[ v = -\frac{1}{(t + 5)^{2}} \]

### Step 3: Differentiate the velocity to find acceleration
Now, we differentiate the velocity \( v \) to find the acceleration \( a \):
\[ a = \frac{dv}{dt} \]
Differentiating \( v = -\frac{1}{(t + 5)^{2}} \):
\[ a = -\frac{d}{dt} \left( (t + 5)^{-2} \right) \]
Using the power rule again:
\[ a = -(-2)(t + 5)^{-3} \cdot \frac{d}{dt}(t + 5) \]
Again, \( \frac{d}{dt}(t + 5) = 1 \):
\[ a = \frac{2}{(t + 5)^{3}} \]

### Step 4: Relate acceleration to velocity
From the expression for velocity, we have:
\[ v = -\frac{1}{(t + 5)^{2}} \]
Taking the reciprocal gives:
\[ (t + 5)^{2} = -\frac{1}{v} \]
Now substituting this into the expression for acceleration:
\[ a = \frac{2}{(t + 5)^{3}} = 2 \left( -\frac{1}{v} \right)^{3/2} \]
This implies:
\[ a \propto v^{-\frac{3}{2}} \]

### Step 5: Final relation
Thus, we can conclude that the acceleration \( a \) is proportional to \( v^{\frac{3}{2}} \):
\[ a \propto v^{\frac{3}{2}} \]