A honeybee is flying parallel to a tabletop at a height h = 2.0 m with a constant velocity v = 20√2

Q: Why can't we just use $s = ut + \frac{1}{2}at^2$ directly with the total distance?

Because the motion is in two dimensions! The velocity is horizontal, but the displacement required is both horizontal and vertical. You must treat it as a vector equation or split it into independent $x$ and $y$ components.

Q: Why does the honey drop move in the bee's frame?

Relative velocity is defined as $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$. Since the drop is stationary ($\vec{v}_{drop} = 0$) and the bee is moving ($\vec{v}_{bee} = \vec{v}$), the velocity of the drop relative to the bee is $0 - \vec{v} = -\vec{v}$.

Q: How do we know the bee uses its maximum acceleration?

The problem asks for the *minimum* time to reach the drop. To cover a specific distance in the shortest possible time starting from rest, you must accelerate as hard as possible.

Q: Why did we get a bi-quadratic equation ($t^4$) instead of a normal quadratic?

Because the acceleration components depend inversely on $t$ and $t^2$. When we square them to find the magnitude $a^2 = a_x^2 + a_y^2$, the $1/t^2$ and $1/t^4$ terms appear. Multiplying by $t^4$ to clear the denominators naturally creates a bi-quadratic equation.

Q: Could the bee have just flown straight down?

If the bee flew straight down, it would still have its initial horizontal velocity $v$. It would drift horizontally and miss the drop! It *must* accelerate horizontally to cancel out its forward momentum while simultaneously accelerating downwards.

Question

A honeybee is flying parallel to a tabletop at a height $h = 2.0$ m with a constant velocity $v = 20\sqrt{2}$ m/s. With its wings, it can achieve a maximum acceleration $a = 400\sqrt{3}$ m/s$^2$. At an instant when the honeybee is vertically above a honey drop on the tabletop, it decides to reach the honey drop. Neglect the reaction time of the honeybee and find the minimum time in which the honeybee can reach the honey drop.

Sigma Percentile · Accepted Answer

### The Setup: A High-Speed Chase in 2D

Imagine you are standing in a room, watching a honeybee. It is flying perfectly horizontally at a height of $h = 2.0 	ext{ m}$, cruising at a constant speed of $v = 20\sqrt{2} 	ext{ m/s}$.

Suddenly, right when it is vertically above a delicious drop of honey on the tabletop, it decides to dive for it.

The bee can accelerate in any direction, but its maximum acceleration is fixed at $a = 400\sqrt{3} 	ext{ m/s}^2$. Our goal is to find the absolute minimum time it takes for the bee to reach the drop.

### The Master Stroke: Shifting the Frame of Reference

We could solve this from the ground frame, tracking the bee's curved parabolic path. But there is a much more elegant way.

Let's jump into the **bee's frame of reference**. By doing this, we make the bee initially at rest, which simplifies our math tremendously.

In the bee's frame, the bee itself feels like it is just hovering in mid-air. But what happens to the honey drop?

The drop appears to be moving backward with a velocity of $-\vec{v}$. So, the drop is not just sitting there; it is sliding away horizontally while being at a depth $h$ below.

### The Kinematic Equation: Forcing an Intercept

To catch the drop, the bee must intercept it. Let's write the position vectors.

The drop's position over time is $\vec{r}_d(t) = -vt\hat{i} - h\hat{j}$.

The bee, starting from rest in this frame, moves purely due to its acceleration, so its position is $\vec{r}_b(t) = \frac{1}{2}\vec{a}t^2$.

For a successful catch, these two position vectors must be exactly equal:
$$ \frac{1}{2}\vec{a}t^2 = -vt\hat{i} - h\hat{j} $$

Now, we want to find the required acceleration. So, let's isolate the acceleration vector $\vec{a}$ by multiplying the entire equation by $2$ and dividing by $t^2$.

This gives us the exact $x$ and $y$ components of the acceleration the bee needs:
$$ \vec{a} = -\frac{2v}{t}\hat{i} - \frac{2h}{t^2}\hat{j} $$

### The Math: Taming the Bi-Quadratic Beast

We know the bee operates at its **maximum acceleration** to minimize the time. So, let's take the magnitude of this acceleration vector.

Using the Pythagorean theorem, the magnitude squared is the sum of the squares of its components:
$$ a^2 = \left(\frac{2v}{t}ight)^2 + \left(\frac{2h}{t^2}ight)^2 = \frac{4v^2}{t^2} + \frac{4h^2}{t^4} $$

This equation looks a bit messy with those denominators. Let's clean it up by multiplying everything by $t^4$.

Bringing all the terms to one side, we get a beautiful quadratic equation, not in $t$, but in $t^2$:
$$ a^2 t^4 - 4v^2 t^2 - 4h^2 = 0 $$

### The Final Execution: Crunching the Numbers

Now for the fun part. Let's bring back the numerical values given in the problem.

We substitute $a = 400\sqrt{3}$, $v = 20\sqrt{2}$, and $h = 2.0$ directly into our bi-quadratic equation:
$$ (400\sqrt{3})^2 t^4 - 4(20\sqrt{2})^2 t^2 - 4(2)^2 = 0 $$

Let's carefully compute the squares. 
$$ 480000 t^4 - 4(800) t^2 - 16 = 0 $$
$$ 480000 t^4 - 3200 t^2 - 16 = 0 $$

Those numbers look a bit intimidating, don't they? But look closely. Every single coefficient is divisible by $16$!

Let's divide the entire equation by $16$ to simplify our lives:
$$ 30000 t^4 - 200 t^2 - 1 = 0 $$

Now we apply the standard quadratic formula, treating $t^2$ as our variable:
$$ t^2 = \frac{200 + \sqrt{(-200)^2 - 4(30000)(-1)}}{2(30000)} $$
$$ t^2 = \frac{200 + \sqrt{40000 + 120000}}{60000} $$

Let's finish the arithmetic. The term inside the square root becomes $160000$, and its square root is a perfect $400$!
$$ t^2 = \frac{200 + 400}{60000} = \frac{600}{60000} = \frac{1}{100} $$

Finally, we take the square root of $\frac{1}{100}$. And we get our answer:
$$ t = 0.1 	ext{ s} $$

In just a tenth of a second, the honeybee calculates its trajectory, accelerates at maximum capacity, and perfectly intercepts the honey drop. Physics in nature is truly amazing!

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$C h oos in g t h e R i g h t F r am e$

$R e l a t i v e M o t i o n S e t u p$

$K in e ma t i cs in R e l a t i v e F r am e$

$I so l a t in g t h e A cce l er a t i o n V ec t or$

$M a g ni t u d e o f A cce l er a t i o n$

$F or min g t h e P o l y n o mia l$

$R a w S u b s t i t u t i o n$

$S q u a r in g t h e T er m s$

$S im pl i f y in g t h e E q u a t i o n$

$A ppl y in g t h e Q u a d r a t i c F or m u l a$

$S o l v in g f or t^{2}$

$F ina l A n s w er$

$T h e W a y F or w a r d$

The Sigma Insight: Relative Velocity

The Setup

The Master Stroke

The Kinematic Equation

The Math

The Final Execution

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Comprehension Passage

When and for how long can the particle C fulfil the given condition?

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At the instant, when the line joining locations of A and B is perpendicular to the line joining locations of B and C, what are the magnitudes of velocities of C relative to A and B respectively?

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Choosing the Right Frame

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relative Motion Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kinematics in Relative Frame

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating the Acceleration Vector

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnitude of Acceleration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forming the Polynomial

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Raw Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Squaring the Terms

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Quadratic Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for t2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward