A body starts with initial velocity u and moves with uniform accelelration . When the velocity has increased to 5u, the acceleration is reversed in direction, the magnitude remaining constant. Find its velocity when it returns to the strating point?

Options

A. `-7 u`
B. `-5u`
C. `-2u`
D. `2u`

Answer

To solve the problem step by step, we will analyze the motion of the body in two phases: the first phase when the body accelerates to a velocity of \(5u\), and the second phase when it decelerates back to the starting point.

### Step 1: Analyze the first phase of motion
In the first phase, the body starts with an initial velocity \(u\) and accelerates uniformly to a final velocity of \(5u\). We can use the equation of motion:

\[ v^2 = u^2 + 2as \]

Where:
- \(v = 5u\) (final velocity)
- \(u = u\) (initial velocity)
- \(a\) = acceleration
- \(s\) = distance traveled

Substituting the known values into the equation:

\[ (5u)^2 = u^2 + 2as \]

This simplifies to:

\[ 25u^2 = u^2 + 2as \]

### Step 2: Rearranging the equation
Now, we can rearrange the equation to find \(2as\):

\[ 25u^2 - u^2 = 2as \]

This gives us:

\[ 24u^2 = 2as \]

Dividing both sides by 2:

\[ 12u^2 = as \quad \text{(Equation 1)} \]

### Step 3: Analyze the second phase of motion
In the second phase, the acceleration is reversed (but has the same magnitude), and the body decelerates back to the starting point. The initial velocity for this phase is \(5u\), and we need to find the final velocity when it returns to the starting point (which we will denote as \(v\)).

Using the same equation of motion:

\[ v^2 = u^2 + 2as \]

In this case:
- Initial velocity \(u = 5u\)
- Final velocity \(v\) (unknown)
- Acceleration \(a\) is now negative, so we write it as \(-a\)
- Distance \(s\) remains the same

Substituting into the equation:

\[ v^2 = (5u)^2 - 2as \]

This simplifies to:

\[ v^2 = 25u^2 - 2as \]

### Step 4: Substitute \(2as\) from Equation 1
From Equation 1, we know that \(2as = 24u^2\). Substituting this into the equation gives:

\[ v^2 = 25u^2 - 24u^2 \]

This simplifies to:

\[ v^2 = u^2 \]

### Step 5: Solve for \(v\)
Taking the square root of both sides, we find:

\[ v = u \]

### Conclusion
Thus, the velocity of the body when it returns to the starting point is \(u\).