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

Options

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

Answer

To solve the problem step by step, we need to find the acceleration of a particle whose position \( s \) is given by the equation \( s = \sqrt{t} \).

### Step 1: Find the velocity
The velocity \( v \) is the first derivative of the position \( s \) with respect to time \( t \):
\[ v = \frac{ds}{dt} \]
Given \( s = \sqrt{t} = t^{1/2} \), we differentiate:
\[ v = \frac{d}{dt}(t^{1/2}) = \frac{1}{2}t^{-1/2} \]
Thus, the velocity is:
\[ v = \frac{1}{2\sqrt{t}} \]

### Step 2: Find the acceleration
The acceleration \( a \) is the derivative of the velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} \]
We need to differentiate \( v = \frac{1}{2}t^{-1/2} \):
\[ a = \frac{d}{dt}\left(\frac{1}{2}t^{-1/2}\right) = \frac{1}{2} \cdot \left(-\frac{1}{2}t^{-3/2}\right) = -\frac{1}{4}t^{-3/2} \]

### Step 3: Relate acceleration to velocity
From our previous steps, we have:
- Velocity \( v = \frac{1}{2}t^{-1/2} \)
- Acceleration \( a = -\frac{1}{4}t^{-3/2} \)

Now, we can express \( t \) in terms of \( v \):
\[ v = \frac{1}{2}t^{-1/2} \implies t^{-1/2} = 2v \implies t = \frac{1}{(2v)^2} = \frac{1}{4v^2} \]

Substituting \( t \) back into the expression for acceleration:
\[ a = -\frac{1}{4}\left(\frac{1}{4v^2}\right)^{-3/2} = -\frac{1}{4} \cdot \left(4v^2\right)^{3/2} \]
Calculating further:
\[ = -\frac{1}{4} \cdot 8v^3 = -2v^3 \]

### Conclusion
Thus, the acceleration \( a \) is proportional to \( v^3 \):
\[ a \propto v^3 \]