A body of mass $2kg$ acted upon by a constant force, travels a distance of $3metres$ in the first second and a further distance of $1metre$ in the next second. The force acting on the body is

The correct option is A 4 Newtons
Given, m=2 kg

Using Newton's equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

In first second, $s_1$ =3 meters distance is covered,

so,

$s_1=u\left(1\right)+\frac{1}{2}a(1{)}^2$

$s_1=u+\frac{1}{2}a$

solving,

$2u+a=6$...........................(1)

In next second, a distance of 1m is covered,

so,

$s_2=s_1+1$

$u\left(2\right)+\frac{1}{2}a(2{)}^2=3+1=4$

solving,

$2u+a=6$.........................(2)

Solving equations (1) and (2),

$a=-2m/s^2$ $u=4m/s$

So, Force$=ma$

$=2\times (-2)=-4N$