A body of mass `3 kg` is moving alone a straight line with a velocity of `24ms^(-1)`.When it is at a point `P` a force of `9 N` acts on the body in a direction opposite to its motion.The time after which it will be at `P` again is.

Options

A. `8s`
B. `16s`
C. `12s`
D. `24s`

Explanation

To solve the problem step by step, we will analyze the motion of the body under the influence of the force acting opposite to its direction of motion.

### Step 1: Identify the given data
- Mass of the body, \( m = 3 \, \text{kg} \)
- Initial velocity, \( u = 24 \, \text{m/s} \)
- Force acting on the body, \( F = 9 \, \text{N} \)

### Step 2: Calculate the acceleration
Using Newton's second law, we can find the acceleration \( a \) caused by the force:
\[ F = m \cdot a \implies a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{9 \, \text{N}}{3 \, \text{kg}} = 3 \, \text{m/s}^2 \]
Since the force is acting in the opposite direction to the motion, the acceleration will be negative:
\[ a = -3 \, \text{m/s}^2 \]

### Step 3: Determine the time to come to a stop
We need to find the time \( t_1 \) it takes for the body to come to a stop (final velocity \( v = 0 \)):
Using the equation of motion:
\[ v = u + a \cdot t \]
Setting \( v = 0 \):
\[ 0 = 24 \, \text{m/s} - 3 \, \text{m/s}^2 \cdot t_1 \]
Solving for \( t_1 \):
\[ 3 \, \text{m/s}^2 \cdot t_1 = 24 \, \text{m/s} \implies t_1 = \frac{24}{3} = 8 \, \text{s} \]

### Step 4: Determine the distance traveled before stopping
Now we calculate the distance \( S_1 \) traveled before the body comes to a stop using the equation:
\[ S = u \cdot t + \frac{1}{2} a t^2 \]
Substituting the known values:
\[ S_1 = 24 \cdot 8 + \frac{1}{2} \cdot (-3) \cdot (8^2) \]
Calculating each term:
\[ S_1 = 192 - \frac{1}{2} \cdot 3 \cdot 64 = 192 - 96 = 96 \, \text{m} \]

### Step 5: Determine the time to return to point P
After coming to a stop, the body will start moving back towards point P with the same acceleration. The time \( t_2 \) to return to point P can be found using the same equations of motion. The initial velocity for this motion is \( 0 \) and the distance to cover is \( 96 \, \text{m} \):
Using the equation:
\[ S = u \cdot t + \frac{1}{2} a t^2 \]
Setting \( u = 0 \):
\[ 96 = 0 + \frac{1}{2} \cdot 3 \cdot t_2^2 \]
Solving for \( t_2 \):
\[ 96 = \frac{3}{2} t_2^2 \implies t_2^2 = \frac{96 \cdot 2}{3} = 64 \implies t_2 = 8 \, \text{s} \]

### Step 6: Calculate the total time
The total time \( T \) taken to return to point P is:
\[ T = t_1 + t_2 = 8 \, \text{s} + 8 \, \text{s} = 16 \, \text{s} \]

### Final Answer
The time after which the body will be at point P again is \( \boxed{16 \, \text{s}} \).