A body of mass $4kg$ is accelerated upon by a constant force, travels a distance of $5m$ in the first second and a distance of $2m$ in the third second. The force acting on the body is:

The correct option is C $6N$

$\text{Step 1: Acceleration calculation}$

As force is constant, hence acceleration is also constant.

Therefore, we can use equations of motion for constant acceleration.

Displacement in $n^{th}$ second is given by

$S_n=u+\frac{a}{2}(2n-1)$

$\text{For First Second,}$ $n=1$

$S_1=u+\frac{a}{2}(2\times 1-1)$

$\Rightarrow$ $5=u+\frac{a}{2}$ $....\left(1\right)$

$\text{For Third second,}$ $n=3$

$S_3=u+\frac{a}{2}(2\times 3-1)$

$\Rightarrow$ $2=u+\frac{5}{2}a$ $....\left(2\right)$

$\text{Step 2: Solving above equations}$

Subtracting equation $\left(2\right)$ from equation $\left(1\right)$

We get, $3=-2a$

$\Rightarrow$ $a=-1.5m/s^2$

Using equation $\left(1\right)$

$5=u-\frac{3}{4}$

$\Rightarrow$ $u=5.75m/s$

$\text{Step 3: Newton's second law on body}$

(Positive Rightwards)

$\sum F=ma$

$\Rightarrow$ $F=4\times (-1.5)N$

$=-6N$

Hence the force acting on the body is $6N$. Option $C$ is correct