Since the cyclist travels along the circumference from point P to point S, which are opposite ends of the diameter of the circle, we can visualize the displacement as the straight-line distance between P and S.
1. Determine the Displacement:
Using the Pythagorean theorem, we find:
\[ \text{Displacement} = R\sqrt{2} = 2\sqrt{2} = \sqrt{8} \, \text{km}. \]
Answer: \(\sqrt{8} \, \text{km}\)
A bead of mass \( m \) slides without friction on the wall of a vertical circular hoop of radius \( R \) as shown in figure. The bead moves under the combined action of gravity and a massless spring \( k \) attached to the bottom of the hoop. The equilibrium length of the spring is \( R \). If the bead is released from the top of the hoop with (negligible) zero initial speed, the velocity of the bead, when the length of spring becomes \( R \), would be (spring constant is \( k \), \( g \) is acceleration due to gravity):
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: