An object is moving velocity `10 ms^(-1)` . A constant force acts for 4 s object and given it a speed of `2 ms^(-1)` in opposite direction. The acceleration produced is

Options

A. `3 ms^(-2)`
B. `-3 ms^(-2)`
C. `6 ms^(-2)`
D. `-6ms^(-2)`

Explanation

To solve the problem, we need to find the acceleration produced when a constant force acts on an object moving in one direction and then causes it to move in the opposite direction.

### Step-by-Step Solution:

1. **Identify the Initial and Final Velocities**:
- The initial velocity \( u = 10 \, \text{m/s} \) (in one direction).
- The final velocity \( v = -2 \, \text{m/s} \) (in the opposite direction).

2. **Use the Equation of Motion**:
The equation of motion we will use is:
\[ v = u + at \]
where:
- \( v \) = final velocity,
- \( u \) = initial velocity,
- \( a \) = acceleration,
- \( t \) = time.

3. **Substitute the Known Values**:
We know:
- \( u = 10 \, \text{m/s} \),
- \( v = -2 \, \text{m/s} \),
- \( t = 4 \, \text{s} \).

Substituting these values into the equation gives:
\[ -2 = 10 + a \cdot 4 \]

4. **Rearrange the Equation**:
Rearranging the equation to solve for \( a \):
\[ -2 - 10 = 4a \]
\[ -12 = 4a \]

5. **Solve for Acceleration \( a \)**:
Dividing both sides by 4:
\[ a = \frac{-12}{4} = -3 \, \text{m/s}^2 \]

6. **Interpret the Result**:
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which means it is a retardation.

### Final Answer:
The acceleration produced is \( -3 \, \text{m/s}^2 \).

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