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. `24s`
B. `21s`
C. `1s`
D. `20s`

Explanation

To solve the problem, we need to find the time it takes for the lizard to catch the insect. We can use the equations of motion for both the lizard and the insect.

### Step-by-Step Solution:

1. **Define the Variables:**
- Let the distance traveled by the lizard be \( s_1 \).
- Let the distance traveled by the insect be \( s_2 \).
- The initial distance between the lizard and the insect is \( 21 \) cm.
- The lizard starts from rest with an acceleration \( a = 2 \, \text{cm/s}^2 \).
- The insect moves with a constant speed \( v = 20 \, \text{cm/s} \).

2. **Write the Equation for the Lizard's Motion:**
- Since the lizard starts from rest, its initial velocity \( u = 0 \).
- The distance traveled by the lizard after time \( t \) is given by the equation:
\[ s_1 = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \cdot 2 \cdot t^2 = t^2 \]

3. **Write the Equation for the Insect's Motion:**
- The distance traveled by the insect after time \( t \) is given by:
\[ s_2 = vt = 20t \]

4. **Set Up the Equation for Catching Up:**
- The lizard catches the insect when the distance it has traveled equals the distance traveled by the insect plus the initial distance of \( 21 \) cm:
\[ s_1 = s_2 + 21 \]
- Substituting the expressions for \( s_1 \) and \( s_2 \):
\[ t^2 = 20t + 21 \]

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

6. **Solving 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. **Calculating the Roots:**
- The two possible solutions for \( t \) are:
\[ t = \frac{42}{2} = 21 \quad \text{and} \quad t = \frac{-2}{2} = -1 \]
- Since time cannot be negative, we discard \( t = -1 \).

8. **Final Answer:**
- Therefore, the lizard will catch the insect after \( t = 21 \) seconds.