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Step 1: Identify the Known Quantities
• Initial distance between the lizard and the insect: 21 cm (the lizard is behind)
• Lizard’s acceleration: $2 \text{ cm/s}^2$ (starting from rest)
• Insect’s constant speed: $20 \text{ cm/s}$
Step 2: Write the Distance Equation for the Lizard
Since the lizard starts from rest and accelerates with $a = 2 \text{ cm/s}^2$, the distance it travels in time $t$ is given by:
$D_l = \frac{1}{2}\,a\,t^2 \;=\; \frac{1}{2}\,\times\,2\,\times\,t^2 \;=\; t^2$
Therefore, the lizard’s distance from its starting point after $t$ seconds is $t^2$ cm.
Step 3: Write the Distance Equation for the Insect
The insect moves at a constant speed of $v = 20 \text{ cm/s}$. Thus, the distance it travels in time $t$ is:
$D_i = v\,t = 20\,t$
So the insect’s distance from its starting point after $t$ seconds is $20t$ cm.
Step 4: Set the Condition for Catching Up
When the lizard catches the insect, both have covered exactly enough distance to be at the same position. However, because the lizard was initially 21 cm behind the insect, the lizard must cover 21 cm more distance than the insect has traveled. This gives the condition:
$D_l = D_i + 21$
Substitute $D_l = t^2$ and $D_i = 20t$ into this equation:
$t^2 = 20t + 21$
Step 5: Solve the Quadratic Equation
Rearrange the equation to standard quadratic form:
$t^2 - 20t - 21 = 0$
To find the roots, we can use factorization or the quadratic formula. Solving this yields two roots:
$t = 21 \quad \text{and} \quad t = -1.$
Step 6: Interpret the Solutions
Since time cannot be negative in this context, we discard $t = -1$ and accept:
$t = 21 \text{ s}.$
Therefore, the lizard will catch the insect after 21 seconds.
Correct Answer: 21 s
