A particle starts from the origin with a velocity of `10 m s^(-1)` and moves with a constant acceleration till the velocity increases to `50 ms^(-1)`. At that instant, the acceleration is suddenly reversed. What will be the velocity of the particle, when it returne to the starticng point?

Options

A. Zero
B. `10ms^(-1)`
C. `50 ms^(-1)`
D. `70 ms^(-1)`

Explanation

To solve the problem step by step, we will break it down into two phases: the first phase where the particle accelerates from 10 m/s to 50 m/s, and the second phase where the particle decelerates back to the starting point.

### Step 1: Determine the acceleration during the first phase

We know:
- Initial velocity (u) = 10 m/s
- Final velocity (v) = 50 m/s

Using the equation of motion:
\[ v^2 = u^2 + 2as_1 \]
where \( s_1 \) is the distance traveled during the first phase and \( a \) is the acceleration.

Rearranging the equation gives:
\[ a = \frac{v^2 - u^2}{2s_1} \]

### Step 2: Calculate the distance traveled in the first phase

Substituting the values into the equation:
\[ 50^2 = 10^2 + 2a s_1 \]
\[ 2500 = 100 + 2a s_1 \]
\[ 2400 = 2a s_1 \]
\[ s_1 = \frac{2400}{2a} = \frac{1200}{a} \]

### Step 3: Determine the distance during the second phase

In the second phase, the particle starts with an initial velocity of 50 m/s and comes to a stop (final velocity = 0 m/s). Using the same equation of motion:
\[ 0 = 50^2 + 2(-a)s_2 \]
where \( s_2 \) is the distance traveled during the second phase.

Rearranging gives:
\[ 0 = 2500 - 2as_2 \]
\[ 2as_2 = 2500 \]
\[ s_2 = \frac{2500}{2a} = \frac{1250}{a} \]

### Step 4: Calculate the total distance traveled

The total distance traveled (s) when the particle returns to the starting point is:
\[ s = s_1 + s_2 \]
Substituting the values we found:
\[ s = \frac{1200}{a} + \frac{1250}{a} = \frac{2450}{a} \]

### Step 5: Find the velocity when returning to the starting point

Now, we need to find the velocity when the particle returns to the starting point. The total distance traveled is \( \frac{2450}{a} \).

Using the equation of motion again:
\[ v^2 = u^2 + 2as \]
Here, \( u = 0 \) (when it returns to the starting point), and \( s = \frac{2450}{a} \):
\[ v^2 = 0 + 2a \left(\frac{2450}{a}\right) \]
\[ v^2 = 2 \cdot 2450 \]
\[ v^2 = 4900 \]
\[ v = \sqrt{4900} = 70 \text{ m/s} \]

### Final Answer
The velocity of the particle when it returns to the starting point is **70 m/s**.