A particle having a velocity `v = v_0` at `t= 0` is decelerated at the rate `|a| = alpha sqrtv ,` where `alpha` is a positive constant.

Options

A. The particle comes to rest at `t = (2sqrtv_0)/alpha`
B. The particle will come to rest at infinity
C. The distance travelled by the particle before coming to rest is `(2v_0^(3//2))/ alpha`
D. The distance travelled by the particle before coming to rest is `(2v_0^(3//2))/(3 alpha)`

Explanation

To solve the problem step by step, we will analyze the motion of a particle that is decelerated at a rate proportional to the square root of its velocity.

### Step 1: Understand the given information
- Initial velocity at time \( t = 0 \) is \( v_0 \).
- The deceleration (negative acceleration) is given by \( |a| = \alpha \sqrt{v} \), where \( \alpha \) is a positive constant.

### Step 2: Set up the equation for acceleration
Since the particle is decelerating, we can express the acceleration as:
\[ a = -\alpha \sqrt{v} \]
Using the relationship between acceleration, velocity, and time, we can write:
\[ \frac{dv}{dt} = -\alpha \sqrt{v} \]

### Step 3: Separate variables for integration
We can rearrange the equation to separate the variables:
\[ \frac{dv}{\sqrt{v}} = -\alpha dt \]

### Step 4: Integrate both sides
Now, we will integrate both sides. The left side will be integrated with respect to \( v \) from \( v_0 \) to \( 0 \) (when the particle comes to rest), and the right side will be integrated with respect to \( t \) from \( 0 \) to \( t \):
\[ \int_{v_0}^{0} \frac{dv}{\sqrt{v}} = -\alpha \int_{0}^{t} dt \]
The integral of \( \frac{1}{\sqrt{v}} \) is \( 2\sqrt{v} \):
\[ 2\sqrt{v} \bigg|_{v_0}^{0} = -\alpha t \]
Calculating the left side gives:
\[ 2(0 - \sqrt{v_0}) = -\alpha t \implies -2\sqrt{v_0} = -\alpha t \implies t = \frac{2\sqrt{v_0}}{\alpha} \]

### Step 5: Find the distance traveled before coming to rest
Next, we need to find the distance \( s \) traveled by the particle before it comes to rest. We can use the relationship:
\[ v \frac{dv}{ds} = -\alpha \sqrt{v} \]
Rearranging gives:
\[ v dv = -\alpha \sqrt{v} ds \]
Dividing both sides by \( \sqrt{v} \):
\[ \sqrt{v} dv = -\alpha ds \]
Integrating both sides, we have:
\[ \int_{v_0}^{0} \sqrt{v} dv = -\alpha \int_{0}^{s} ds \]
The integral of \( \sqrt{v} \) is \( \frac{2}{3} v^{3/2} \):
\[ \frac{2}{3} v^{3/2} \bigg|_{v_0}^{0} = -\alpha s \]
Calculating gives:
\[ \frac{2}{3} (0 - v_0^{3/2}) = -\alpha s \implies -\frac{2}{3} v_0^{3/2} = -\alpha s \implies s = \frac{2}{3\alpha} v_0^{3/2} \]

### Final Answers
- The time taken to come to rest is:
\[ t = \frac{2\sqrt{v_0}}{\alpha} \]
- The distance traveled before coming to rest is:
\[ s = \frac{2}{3\alpha} v_0^{3/2} \]