A lizard, at an initial distance of 21 cm behind an insect, moves from rest with an acceleration of `2cms^(-2)` and pursues the insect which is crawling uniformly along a straight line at a speed of `20cms^(-1)`. Then the lizard will catch the insect after

Options

A. (a) 24 s
B. (b) 21 s
C. (c) 1 s
D. (d) 20 s

Answer

To solve the problem, we need to find out how long it takes for the lizard to catch the insect. We will use the equations of motion for both the lizard and the insect.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Initial distance between the lizard and the insect, \( d = 21 \, \text{cm} \)
- Acceleration of the lizard, \( a_L = 2 \, \text{cm/s}^2 \)
- Speed of the insect, \( v_I = 20 \, \text{cm/s} \)

2. **Write the Equation of Motion for the Lizard:**
The lizard starts from rest, so its initial velocity \( u_L = 0 \). The distance covered by the lizard after time \( t \) is given by:
\[ S_L = u_L t + \frac{1}{2} a_L t^2 = 0 + \frac{1}{2} (2) t^2 = t^2 \]
Thus, the distance covered by the lizard is:
\[ S_L = t^2 \]

3. **Write the Equation of Motion for the Insect:**
The insect is moving at a constant speed, so its distance covered after time \( t \) is:
\[ S_I = v_I t = 20t \]

4. **Set Up the Equation:**
Since the lizard is initially 21 cm behind the insect, the distance covered by the lizard must equal the distance covered by the insect plus the initial distance:
\[ S_L = S_I + 21 \]
Substituting the expressions for \( S_L \) and \( S_I \):
\[ t^2 = 20t + 21 \]

5. **Rearrange the Equation:**
Rearranging gives us a standard quadratic equation:
\[ t^2 - 20t - 21 = 0 \]

6. **Solve the Quadratic Equation:**
We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -20, c = -21 \):
\[ t = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot (-21)}}{2 \cdot 1} \]
\[ t = \frac{20 \pm \sqrt{400 + 84}}{2} \]
\[ t = \frac{20 \pm \sqrt{484}}{2} \]
\[ t = \frac{20 \pm 22}{2} \]

7. **Calculate the Possible Values for \( t \):**
- Positive root:
\[ t = \frac{42}{2} = 21 \, \text{s} \]
- Negative root:
\[ t = \frac{-2}{2} = -1 \, \text{s} \quad (\text{not valid since time cannot be negative}) \]

8. **Conclusion:**
The lizard will catch the insect after \( t = 21 \, \text{s} \).

### Final Answer:
The lizard will catch the insect after **21 seconds**.