Step 1: Setup and Frame of Reference
The honeybee moves with a constant velocity $v$ at height $h$. It decides to reach a stationary honey drop located vertically below it at $t=0$. The bee can accelerate with a maximum acceleration $a$.
Let us define the ground frame coordinates:
Step 2: Kinematics Equation
Let the bee reach the drop at time $t$. We need to find the minimum $t$.
The final position of the bee must be $(0, 0)$.
Using the equation of motion $\vec{r}(t) = \vec{r}_0 + \vec{u}t + \frac{1}{2}\vec{a}t^2$:
X-axis motion:
$$ 0 = 0 + vt + \frac{1}{2} a_x t^2 \implies a_x = -\frac{2v}{t} $$
Y-axis motion:
$$ 0 = h + 0\cdot t + \frac{1}{2} a_y t^2 \implies a_y = -\frac{2h}{t^2} $$
The magnitude of the required acceleration is:
$$ a_{req} = \sqrt{a_x^2 + a_y^2} = \sqrt{\left(-\frac{2v}{t}\right)^2 + \left(-\frac{2h}{t^2}\right)^2} $$
To minimize the time $t$, the bee should use its maximum available acceleration $a$. Thus, we set $a_{req} = a$:
$$ a^2 = \frac{4v^2}{t^2} + \frac{4h^2}{t^4} $$
Step 3: Solving for Time $t$
Multiply the equation by $t^4$ to form a quadratic in $t^2$:
$$ a^2 t^4 = 4v^2 t^2 + 4h^2 $$
$$ a^2 (t^2)^2 – 4v^2 (t^2) – 4h^2 = 0 $$
Let $T = t^2$. The equation becomes:
$$ a^2 T^2 – 4v^2 T – 4h^2 = 0 $$
Solving for $T$ using the quadratic formula:
$$ T = \frac{4v^2 \pm \sqrt{16v^4 – 4(a^2)(-4h^2)}}{2a^2} $$
$$ T = \frac{4v^2 + \sqrt{16v^4 + 16a^2 h^2}}{2a^2} $$
(We discard the negative root since $T=t^2$ must be positive).
$$ T = \frac{2(v^2 + \sqrt{v^4 + a^2 h^2})}{a^2} $$
Since $t = \sqrt{T}$:
$$ t = \frac{\sqrt{2(v^2 + \sqrt{v^4 + a^2 h^2})}}{a} $$
Step 4: Numerical Substitution
Given values:
Derivation using Relative Frame:
“It is better to use a frame moving with the initial velocity of the honeybee.”
1. Define the Frame:
In this frame, the Honeybee’s initial velocity is $0$.
The Honey Drop (initially at rest) has a relative velocity $\vec{v}_{rel} = \vec{v}_{drop} – \vec{v}_{bee} = 0 – \vec{v} = -\vec{v}$.
So, the drop appears to move horizontally backwards with speed $v$.
Initial position of drop: Vertical height $h$ above/below bee (depending on setup, distance is $h$).
2. Interception Condition:
The Bee accelerates with magnitude $a$ from rest to intercept the moving drop.
Let interception time be $t$.
Position of Drop at $t$: $x_D = vt$ (horizontal), $y_D = h$ (vertical distance).
Distance to Drop: $D(t) = \sqrt{(vt)^2 + h^2}$.
The Bee must cover this distance with acceleration $a$.
$$ \frac{1}{2} a t^2 = \sqrt{v^2 t^2 + h^2} $$
3. Solving for Minimum Time:
Square both sides:
$$ \frac{1}{4} a^2 t^4 = v^2 t^2 + h^2 $$
Rearrange into a quadratic equation for $T = t^2$:
$$ \frac{a^2}{4} (t^2)^2 – v^2 (t^2) – h^2 = 0 $$
Solve for $t^2$ using the quadratic formula:
$$ t^2 = \frac{v^2 + \sqrt{(v^2)^2 – 4(\frac{a^2}{4})(-h^2)}}{2(\frac{a^2}{4})} $$
$$ t^2 = \frac{v^2 + \sqrt{v^4 + a^2 h^2}}{a^2/2} = \frac{2(v^2 + \sqrt{v^4 + a^2 h^2})}{a^2} $$
Taking the square root:
$$ t = \frac{\sqrt{2(v^2 + \sqrt{v^4 + h^2 a^2})}}{a} $$
Calculation:
$v = 20\sqrt{2}$, $h = 2.0$, $a = 400\sqrt{3}$.
$v^2 = 800$, $v^4 = 640000$.
$a^2 h^2 = (480000)(4) = 1920000$.
$\sqrt{640000 + 1920000} = \sqrt{2560000} = 1600$.
Numerator term: $2(800 + 1600) = 4800$.
$t = \frac{\sqrt{4800}}{400\sqrt{3}} = \frac{40\sqrt{3}}{400\sqrt{3}} = 0.1 \text{ s}$.
