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

Options

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

Answer

To solve the problem, we need to find the acceleration of the particle based on the given position-time relationship \( x = (t + 5)^{-1} \). We will follow these steps:

### Step 1: Differentiate the position function to find velocity

Given:
\[ x = (t + 5)^{-1} \]

To find the velocity \( v \), we differentiate \( x \) with respect to \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt} \left( (t + 5)^{-1} \right) \]

Using the chain rule:
\[ v = -1 \cdot (t + 5)^{-2} \cdot \frac{d}{dt}(t + 5) = -\frac{1}{(t + 5)^2} \]

### Step 2: Differentiate the velocity function to find acceleration

Now, we differentiate the velocity \( v \) to find the acceleration \( a \):
\[ a = \frac{dv}{dt} = \frac{d}{dt} \left( -\frac{1}{(t + 5)^2} \right) \]

Using the chain rule again:
\[ a = 2 \cdot (t + 5)^{-3} \cdot \frac{d}{dt}(t + 5) = \frac{2}{(t + 5)^3} \]

### Step 3: Express acceleration in terms of velocity

From the velocity equation, we have:
\[ v = -\frac{1}{(t + 5)^2} \implies (t + 5)^2 = -\frac{1}{v} \]

Substituting this into the acceleration equation:
\[ a = \frac{2}{(t + 5)^3} = 2 \cdot \left(-\frac{1}{v}\right)^{3/2} = 2 \cdot \frac{(-1)^{3/2}}{v^{3/2}} \]

Since we are interested in the magnitude of acceleration:
\[ |a| = 2 \cdot \frac{1}{|v|^{3/2}} \]

### Step 4: Conclusion about proportionality

From the expression we derived, we can see that the acceleration \( a \) is proportional to \( |v|^{-3/2} \). Thus, we can conclude:
\[ a \propto |v|^{3/2} \]

### Final Answer

The acceleration of the particle is proportional to \( v^{3/2} \).

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