1. Let the insect start at position $x=0$ at $t=0$.
   Then the lizard is initially $21$ cm behind it, so its initial position is $x=-21$ cm.

2. Position of the insect
   The insect moves uniformly with speed $20\ \text{cm/s}$, so after time $t$ seconds: $x_I = 20t$

3. Position of the lizard
   The lizard starts from rest with acceleration $2\ \text{cm/s}^2$.
   Hence displacement in time $t$ is: $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2)t^2 = t^2$ Since it starts from $x=-21$ cm, its position is: $x_L = -21 + t^2$

4. Condition for catching the insect
   The lizard catches the insect when both positions are equal: $-21 + t^2 = 20t$ $t^2 - 20t - 21 = 0$

5. Solve the quadratic equation $t^2 - 20t - 21 = 0$ $t = \frac{20 \pm \sqrt{400 + 84}}{2} = \frac{20 \pm \sqrt{484}}{2} = \frac{20 \pm 22}{2}$ So, $t = 21 \quad \text{or} \quad t = -1$ Negative time is not physically meaningful, so: $t=21\ \text{s}$

6. Check options

   • A: $20$ s ❌
   • B: $1$ s ❌
   • C: $21$ s ✅
   • D: $24$ s ❌

Therefore, the lizard catches the insect after $21$ s.