1. Given data

   • Force: $F(t)=6t$
   • Mass: $m=1\,\text{kg}$
   • Initial velocity: $u=0$
   • Time interval: first $1\,\text{s}$

2. Find acceleration as a function of time

   Using Newton’s second law, $F=ma$ $6t = 1\cdot a$ $a(t)=6t$

3. Find velocity as a function of time

   Since $a=\frac{dv}{dt}=6t$ integrate with respect to $t$: $v=\int 6t\,dt = 3t^2 + C$

   The particle starts from rest, so at $t=0$, $v=0$. Hence $C=0$.

   Therefore, $v(t)=3t^2$

4. Find velocity at $t=1\,\text{s}$

   $v(1)=3(1)^2=3\,\text{m/s}$

5. Use work-energy theorem

   Work done in first $1$ second is equal to change in kinetic energy: $W=\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

   Since $u=0$, $W=\frac{1}{2}(1)(3)^2$ $W=\frac{9}{2}=4.5\,\text{J}$

6. Check options

   • A: $18\,\text{J}$ ❌
   • B: $4.5\,\text{J}$ ✅
   • C: $22\,\text{J}$ ❌
   • D: $9\,\text{J}$ ❌

Therefore, the correct option is B.