A lizard, at an initial distance of 21 cm behind an insect, moves from rest with an acceleration of

Q: Wait, why can't we just use relative velocity here?

That is a fantastic question! We can only use the simple formula $v_{rel} = \frac{d}{t}$ when the relative acceleration is zero. Here, the lizard is accelerating while the insect is at constant velocity, meaning the relative acceleration is non-zero. Because the velocity of the lizard is changing, the relative velocity is not constant, so we must use the equations of motion instead.

Q: Why do we reject the negative root $t = -1$?

In physics, time $t$ represents the duration elapsed since the start of the observation (the moment the lizard begins its pursuit). A negative value for time would imply a state before the chase even began, which is physically meaningless in this context. We only care about the future, so we discard the negative root.

Q: Is there a way to solve this using relative acceleration?

Yes, absolutely! You can define the relative position $s_{rel} = s_L - s_I$. The initial relative position is $-21 \text{ cm}$. The relative acceleration is $2 \text{ cm/s}^2$, and the initial relative velocity is $-20 \text{ cm/s}$. Using $s_{rel} = u_{rel}t + \frac{1}{2}a_{rel}t^2$, you get $-21 = -20t + \frac{1}{2}(2)t^2$, which simplifies to the same quadratic equation. It is a great way to verify your answer!

Question

A lizard, at an initial distance of 21 cm behind an insect, moves from rest with an acceleration of $2 	ext{ cm/s}^2$ and pursues the insect which is crawling uniformly along a straight line at a speed of 20 cm/s. Then the lizard will catch the insect after

Options:

(A) 20 s

(B) 1 s

(C) 21 s

(D) 24 s

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To solve this, we must first establish our frame of reference. Let us place the lizard at the origin of our coordinate system, $x = 0$, at the moment the chase begins ($t = 0$).

Since the insect is $21 	ext{ cm}$ ahead, its initial position is $x = 21 	ext{ cm}$.

For the lizard, the initial velocity is $u_L = 0$, and its acceleration is $a_L = 2 	ext{ cm/s}^2$. For the insect, the velocity is constant, so $v_I = 20 	ext{ cm/s}$, and its acceleration is $a_I = 0$.

### The Equations of Motion

Now, we describe the position of each participant as a function of time $t$. For the lizard, we use the second equation of motion, which relates displacement to time under constant acceleration:

$$s_L = u_Lt + \frac{1}{2}a_Lt^2$$

Since the lizard starts from rest, $u_L = 0$. Substituting the acceleration $a_L = 2 	ext{ cm/s}^2$, the equation simplifies beautifully:

$$s_L = 0 \cdot t + \frac{1}{2}(2)t^2 = t^2$$

This tells us that the lizard's distance from the origin grows quadratically with time. Now, consider the insect. Since it moves with constant velocity, its displacement is simply $s_{insect} = 20t$.

However, we must remember that the insect started $21 	ext{ cm}$ ahead. Therefore, its total position from the origin is:

$$s_I = 21 + 20t$$

### The Catch Condition

The moment of the catch is the moment when both the lizard and the insect occupy the exact same position on the path. Mathematically, this means their position functions must be equal:

$$s_L = s_I$$

Substituting our expressions, we arrive at our master equation:

$$t^2 = 21 + 20t$$

### The Quadratic Resolution

To solve for $t$, we rearrange this into the standard form of a quadratic equation, $At^2 + Bt + C = 0$:

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

We need to find the roots of this equation. We are looking for two numbers that multiply to $-21$ and add up to $-20$. Those numbers are $-21$ and $1$.

Thus, we can factor the equation:

$$(t - 21)(t + 1) = 0$$

This gives us two potential solutions: $t = 21 	ext{ s}$ or $t = -1 	ext{ s}$.

### Final Physical Reality

In the world of pure algebra, both solutions are valid. But in the world of physics, we must apply our intuition. Time $t$ represents the duration of the chase, which must be a positive value.

A time of $t = -1 	ext{ s}$ would imply the catch happened before the chase even started, which is physically impossible. Therefore, we discard the negative root.

We are left with **$t = 21 	ext{ s}$**. After twenty-one seconds of accelerating, the lizard finally closes the gap and catches the insect.

Select Answer:

Visualized Solution

Visualizing the Chase

Lizard's Equation of Motion

Insect's Position

The Catch Condition

Forming the Quadratic Equation

Splitting the Middle Term

Solving for Roots

Final Conclusion

The Sigma Insight: Solution of Quadratic Equations

Analyzing the Setup

The Equations of Motion

The Catch Condition

The Quadratic Resolution

Final Physical Reality

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Select Answer:

Visualized Solution

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