A body starting with a velocity 'v' returns to its initial position after 't' second with the same speed, along the same line. Acceleration of the particle is

Options

A. `(-2v)/t`
B. `(2v)/t`
C. `v/(2t)`
D. `t/(2v)`

Answer

To solve the problem step by step, we will use the equations of motion.

### Step 1: Understand the problem
A body starts with an initial velocity \( v \) and returns to its initial position after \( t \) seconds with the same speed \( v \) but in the opposite direction. This means that the final velocity is \( -v \).

### Step 2: Identify the known values
- Initial velocity \( u = v \)
- Final velocity \( v_f = -v \)
- Time \( t \)

### Step 3: Use the first equation of motion
The first equation of motion is given by:
\[ v_f = u + at \]
Where:
- \( v_f \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration
- \( t \) = time

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:
\[ -v = v + at \]

### Step 5: Rearrange the equation to solve for acceleration \( a \)
Rearranging the equation gives:
\[ -v - v = at \]
\[ -2v = at \]

### Step 6: Solve for acceleration \( a \)
Now, divide both sides by \( t \):
\[ a = \frac{-2v}{t} \]

### Final Result
The acceleration of the particle is:
\[ a = -\frac{2v}{t} \]

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