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 use the equations of motion.

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

The particle starts from the origin with an initial velocity \( U = 10 \, \text{m/s} \) and accelerates to a final velocity \( V = 50 \, \text{m/s} \). We can use the third equation of motion:

\[ V^2 = U^2 + 2a x \]

Where:
- \( V = 50 \, \text{m/s} \)
- \( U = 10 \, \text{m/s} \)
- \( a \) is the acceleration
- \( x \) is the displacement during this phase

Substituting the known values:

\[ (50)^2 = (10)^2 + 2a x \]

Calculating the squares:

\[ 2500 = 100 + 2a x \]

Rearranging gives:

\[ 2a x = 2500 - 100 \]
\[ 2a x = 2400 \]
\[ a x = 1200 \quad \text{(1)} \]

### Step 2: Determine the final velocity after the acceleration is reversed

Now, when the acceleration is suddenly reversed, the initial velocity for this phase is \( U = 50 \, \text{m/s} \). We want to find the final velocity \( V \) when the particle returns to the starting point (displacement \( -x \)).

Using the third equation of motion again:

\[ V^2 = U^2 + 2(-a)(-x) \]

Substituting the known values:

\[ V^2 = (50)^2 + 2(-a)(-x) \]
\[ V^2 = 2500 + 2ax \]

From equation (1), we know \( ax = 1200 \):

\[ V^2 = 2500 + 2(1200) \]
\[ V^2 = 2500 + 2400 \]
\[ V^2 = 4900 \]

### Step 3: Calculate the final velocity

Taking the square root to find \( V \):

\[ V = \sqrt{4900} \]
\[ V = 70 \, \text{m/s} \]

Thus, the velocity of the particle when it returns to the starting point is \( 70 \, \text{m/s} \).

### Final Answer
The final velocity of the particle when it returns to the starting point is \( 70 \, \text{m/s} \).

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