Question

Two beetles, A and B, hold the ends of a slightly stretched rubber band laid out on a horizontal table top. The initial positions of the beetles are (0, 4l) and (0, -l), as shown in the diagram. A knot, tied on the band, is initially located at the origin. The beetles begin to run simultaneously along the tabletop. Beetle A starts from rest in the +y direction at constant unknown acceleration a. Beetle B moves at quickly acquired constant speed v in the +x direction. As the beetles run, the knot on the band passes through a point with coordinates (2l, l). Find the acceleration of beetle A.

Answer 1

The acceleration of beetle A is $\frac{8v^2}{5l}$.

Answer 2

1. Determine Knot's Division Ratio: Calculate the initial distances from the knot (origin) to beetle A ($(0,4l)$) and beetle B ($(0,-l)$). These are $4l$ and $l$ respectively. The knot divides the segment AB in the ratio $AK:KB = 4l:l = 4:1$.

2. Apply Section Formula: Use the section formula for a point dividing a line segment in a given ratio. If the knot $K$ divides $AB$ in the ratio $4:1$, its position vector $P_K(t)$ is given by $P_K(t) = \frac{1 \cdot P_A(t) + 4 \cdot P_B(t)}{1+4} = \frac{P_A(t) + 4 P_B(t)}{5}$.

3. Express Beetle Positions: Write down the position vectors of beetles A and B at time $t$.

   • $P_A(t) = (0, 4l + \frac{1}{2}at^2)$ (initial velocity is 0, constant acceleration $a$ in +y direction)
   • $P_B(t) = (vt, -l)$ (constant velocity $v$ in +x direction)

4. Substitute and Form Knot's Position: Substitute $P_A(t)$ and $P_B(t)$ into the section formula to get the general position of the knot $P_K(t) = \left( \frac{4vt}{5}, \frac{at^2}{10} \right)$.

5. Use Given Knot Position: The knot passes through $(2l, l)$. Equate the components of $P_K(t)$ to $(2l, l)$.

   • $\frac{4vt}{5} = 2l \implies t = \frac{5l}{2v}$
   • $\frac{at^2}{10} = l$

6. Solve for Acceleration: Substitute the expression for $t$ from the x-coordinate equation into the y-coordinate equation and solve for $a$.

   • $\frac{a}{10} \left( \frac{5l}{2v} \right)^2 = l$
   • $\frac{a}{10} \frac{25l^2}{4v^2} = l$
   • $a = \frac{10 \cdot 4v^2 \cdot l}{25l^2} = \frac{40v^2}{25l} = \frac{8v^2}{5l}$