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 need to analyze the motion of the particle under the given conditions.

### Step 1: Understand the given information
The particle has an initial velocity \( v_0 \) at \( t = 0 \) and is decelerated at a rate given by \( |a| = \alpha \sqrt{v} \), where \( \alpha \) is a positive constant. Since it is deceleration, we can express acceleration as:
\[ a = -\alpha \sqrt{v} \]

### Step 2: Relate acceleration to velocity
We know that acceleration \( a \) can also be expressed in terms of velocity \( v \) and time \( t \) as:
\[ a = \frac{dv}{dt} \]
Thus, we can write:
\[ \frac{dv}{dt} = -\alpha \sqrt{v} \]

### Step 3: Separate variables and integrate
We can rearrange the equation to separate the variables:
\[ \frac{dv}{\sqrt{v}} = -\alpha dt \]
Now, we 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 \( t \):
\[ \int_{v_0}^{0} \frac{dv}{\sqrt{v}} = -\alpha \int_{0}^{t} dt \]

### Step 4: Perform the integration
The integral on the left side becomes:
\[ \left[ 2\sqrt{v} \right]_{v_0}^{0} = 2\sqrt{0} - 2\sqrt{v_0} = -2\sqrt{v_0} \]
The integral on the right side is simply:
\[ -\alpha t \]
Setting these equal gives:
\[ -2\sqrt{v_0} = -\alpha t \]
Thus, we find:
\[ t = \frac{2\sqrt{v_0}}{\alpha} \]

### Step 5: Find the distance traveled
To find the distance traveled before the particle comes to rest, we can use the relationship:
\[ v \frac{dv}{ds} = -\alpha \sqrt{v} \]
Rearranging gives:
\[ \frac{v}{\sqrt{v}} dv = -\alpha ds \]
This simplifies to:
\[ \sqrt{v} dv = -\alpha ds \]
Integrating both sides, we have:
\[ \int_{v_0}^{0} \sqrt{v} dv = -\alpha \int_{0}^{s} ds \]

### Step 6: Perform the integration for distance
The left side integral becomes:
\[ \left[ \frac{2}{3} v^{3/2} \right]_{v_0}^{0} = \frac{2}{3} (0 - v_0^{3/2}) = -\frac{2}{3} v_0^{3/2} \]
The right side integral gives:
\[ -\alpha s \]
Setting these equal gives:
\[ -\frac{2}{3} v_0^{3/2} = -\alpha s \]
Thus, we find:
\[ s = \frac{2}{3} \frac{v_0^{3/2}}{\alpha} \]

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