1. Analyzing the Trajectory Geometry:

The stone’s trajectory is a parabola. On the graph paper, the range is $R_{graph} = 1.0$ m and the maximum height is $H_{graph} = 0.25$ m.

For any projectile motion, the relationship between maximum height ($H$) and range ($R$) depends on the projection angle $\theta$:

$$ \frac{H}{R} = \frac{\tan \theta}{4} $$

Substitute the values from the graph:

$$ \frac{0.25}{1.0} = \frac{1}{4} = \frac{\tan \theta}{4} \implies \tan \theta = 1 \implies \theta = 45^\circ $$

2. Equation of the Path on Paper:

The equation of a projectile trajectory is given by $y = x \tan\theta (1 – \frac{x}{R})$.

Using $\tan \theta = 1$ and $R_{graph} = 1$ m, the equation of the curve on the paper (in meters) is:

$$ y = x(1 – x) = x – x^2 $$

3. Radius of Curvature:

The insect moves along this curve with a uniform speed $v_{insect}$. The acceleration of the insect corresponds to the centripetal acceleration, which is maximum when the radius of curvature $\rho$ is minimum.

The radius of curvature formula is:

$$ \rho = \frac{[1 + (dy/dx)^2]^{3/2}}{|d^2y/dx^2|} $$

From $y = x – x^2$:

• First derivative: $y’ = 1 – 2x$
• Second derivative: $y” = -2$

The radius is minimum at the peak where $y’ = 0$ (i.e., at $x = 0.5$).

$$ \rho_{min} = \frac{[1 + 0^2]^{3/2}}{|-2|} = \frac{1}{2} = 0.5 \text{ m} $$

4. Calculation of Maximum Acceleration:

The insect moves with a uniform speed $v_{insect} = 1.0 \text{ cm/s} = 10^{-2} \text{ m/s}$.

Since the tangential acceleration is zero (uniform speed), the total acceleration is just the normal (centripetal) acceleration:

$$ a_{max} = \frac{v_{insect}^2}{\rho_{min}} $$

Substitute the values:

$$ a_{max} = \frac{(10^{-2})^2}{0.5} = \frac{10^{-4}}{0.5} = 2 \times 10^{-4} \text{ m/s}^2 $$