A particle moves in a straight line so that `s=sqrt(t)`, then its acceleration is proportional to

Options

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

Explanation

To solve the problem step by step, we start with the given position function of the particle:

### Step 1: Write down the position function
The position of the particle is given by:
\[ s = \sqrt{t} \]

### Step 2: Differentiate to find velocity
To find the velocity \( v \), we differentiate the position function \( s \) with respect to time \( t \):
\[ v = \frac{ds}{dt} = \frac{d}{dt}(\sqrt{t}) \]
Using the power rule, we can rewrite \( \sqrt{t} \) as \( t^{1/2} \):
\[ v = \frac{1}{2} t^{-1/2} = \frac{1}{2\sqrt{t}} \]

### Step 3: Differentiate to find acceleration
Next, we differentiate the velocity function \( v \) to find the acceleration \( a \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{1}{2\sqrt{t}}\right) \]
Using the power rule again:
\[ a = \frac{1}{2} \cdot (-\frac{1}{2}) t^{-3/2} = -\frac{1}{4} t^{-3/2} \]

### Step 4: Determine proportionality
From the expression for acceleration, we see that:
\[ a = -\frac{1}{4} t^{-3/2} \]
This indicates that the acceleration \( a \) is proportional to \( t^{-3/2} \).

### Final Answer
Thus, the acceleration of the particle is proportional to \( t^{-3/2} \).

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