A body of mass 10 kg is moving with a constant velocity of 10 m/s. When a constant force acts for 4 sec on it, it moves with a velocity 2 m/sec in the opposite direction. The acceleration produced in it is

The correct option is -3 m/s².

Step-by-step Explanation:

To find the acceleration produced in the body, we can use the first equation of motion, which relates initial velocity, final velocity, acceleration, and time.

1. Identify the given values:
- Initial velocity ($u$): The body is initially moving with a velocity of 10 m/s. Let this direction be positive.
Therefore, $u=10\text{m/s}$.
- Final velocity ($v$): After the force acts, the body moves at 2 m/s in the opposite direction. Since the direction is reversed, we assign it a negative sign.
Therefore, $v=-2\text{m/s}$.
- Time interval ($t$): The force acts for a duration of 4 seconds.
Therefore, $t=4\text{s}$.

2. Recall the first equation of motion:

$v=u+at$

Where $a$ represents the acceleration of the body.

3. Substitute the values and solve for acceleration ($a$):

$-2=10+a\left(4\right)$

Subtract 10 from both sides of the equation:

$-2-10=4a$

$-12=4a$

Now, divide both sides by 4 to solve for $a$:

$a=\frac{-12}{4}$

$a=-3{\text{m/s}}^2$

Thus, the acceleration produced in the body is $-3{\text{m/s}}^2$. The negative sign indicates that the acceleration acts in the direction opposite to the body's initial motion.