QJEEM05Ma31d7: JEE Main 2005 Mathematics - Class 12 JEEM05Ma31d7 Solution: A lizard, at an initial distance of 21 cm behind an insect…

The motion of the lizard, which starts from rest and accelerates at a rate of $a = 2 \, \text{cm/s}^2$ , can be described by the equation of motion :

$D_l = \frac{1}{2} a t^2$

where $D_l$ is the distance the lizard travels, $a$ is its acceleration, and $t$ is the time.

The insect, moving at a constant speed of $v = 20 \, \text{cm/s}$ , has a motion that can be simply described as:

$D_i = v t$

where $D_i$ is the distance the insect travels, $v$ is its velocity, and $t$ is the time.

Since the lizard starts 21 cm behind the insect and needs to catch up, it must travel the distance the insect travels plus an additional 21 cm. Equating these two distances gives us :

$\frac{1}{2} a t^2 = v t + 21$

Substituting the given values into this equation gives :

$t^2 = \frac{(v t + 21) \cdot 2}{a}$

Substituting $a = 2 \, \text{cm/s}^2$ and $v = 20 \, \text{cm/s}$ , we simplify to :

$t^2 = 20t + 21$

This simplifies to a quadratic equation :

$t^2 - 20t - 21 = 0$

Solving this quadratic equation for $t$ gives the solutions $t = 21 \, \text{s}$ and $t = -1 \, \text{s}$ . Since time cannot be negative, we discard the -1 s solution. Therefore, the lizard will catch the insect after 21 seconds.

So, the correct answer is Option C : 21 s.