A body starts from a point O with a velocity v and uniform acceleration `alpha. `The direction of acceleration is revarsed when the velocity of the body becomes 5v. The velocity of body at point O will be

Options

A. `-v`
B. `-5v`
C. `-7v`
D. `-9v`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
The body starts from point O with an initial velocity \( v \) and a uniform acceleration \( \alpha \).

### Step 2: Determine the final velocity before acceleration reversal
The problem states that the direction of acceleration is reversed when the velocity of the body becomes \( 5v \).

### Step 3: Use the kinematic equation
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 5v \) (final velocity),
- \( u = v \) (initial velocity),
- \( a = \alpha \) (acceleration),
- \( s \) is the distance traveled.

Substituting these values into the equation gives:
\[ (5v)^2 = v^2 + 2\alpha s \]
This simplifies to:
\[ 25v^2 = v^2 + 2\alpha s \]
\[ 24v^2 = 2\alpha s \]
\[ s = \frac{12v^2}{\alpha} \]

### Step 4: Analyze the motion after acceleration reversal
After reaching the velocity of \( 5v \), the acceleration is reversed. Now, we need to find the new velocity \( v' \) when the body returns to point O.

### Step 5: Apply the kinematic equation again
Now, we apply the kinematic equation again for the return journey:
\[ v'^2 = (5v)^2 + 2(-\alpha)(-s) \]
Here, the acceleration is negative because it is in the opposite direction. Thus, substituting the values:
\[ v'^2 = (5v)^2 + 2\alpha s \]
Substituting \( s = \frac{12v^2}{\alpha} \):
\[ v'^2 = 25v^2 + 2\alpha \left(\frac{12v^2}{\alpha}\right) \]
This simplifies to:
\[ v'^2 = 25v^2 + 24v^2 \]
\[ v'^2 = 49v^2 \]
Taking the square root gives:
\[ v' = \pm 7v \]

### Step 6: Determine the direction of velocity
Since the body is moving back towards point O, the velocity \( v' \) should be negative. Therefore:
\[ v' = -7v \]

### Final Answer
The velocity of the body at point O will be:
\[ \boxed{-7v} \]