Q35: Velocities of C relative to A and B

The question specifies the instant when the line joining A and B is perpendicular to the line joining B and C. Since A and B lie on the horizontal line of motion, the segment $BC$ must be vertical.

Geometry: Let the line AB coincide with the x-axis. $B$ is at $(x_B, 0)$, $C$ is at $(x_B, 30)$. $A$ is at $(x_A, 0)$. $v_A = (5, 0)$, $v_B = (-5, 0)$.

Constraint 1: Distance BC is constant (30m). The relative velocity between B and C along the vertical line BC must be zero. $\vec{r}_{CB} = (0, 30)$. $\vec{v}_C – \vec{v}_B = (u – (-5), v) = (u+5, v)$. Dot product: $(0, 30) \cdot (u+5, v) = 0 \Rightarrow 30v = 0 \Rightarrow v = 0$. So, particle C is moving horizontally.

Constraint 2: Distance AC is constant (40m). $\vec{r}_{CA} = C – A$. Since $BC=30$ and $AC=40$, the triangle is right-angled at B. $AB = \sqrt{40^2 – 30^2} = 10\sqrt{7}$. $C – A = (10\sqrt{7}, 30)$. Relative velocity $\vec{v}_C – \vec{v}_A = (u – 5, 0)$. Dot product: $(10\sqrt{7}, 30) \cdot (u-5, 0) = 0$. $10\sqrt{7}(u-5) = 0 \Rightarrow u = 5$.

Results: $\vec{v}_C = (5, 0) \text{ m/s}$.
Velocity of C relative to A: $\vec{v}_{C/A} = \vec{v}_C – \vec{v}_A = (5,0) – (5,0) = 0 \text{ m/s}$.
Velocity of C relative to B: $\vec{v}_{C/B} = \vec{v}_C – \vec{v}_B = (5,0) – (-5,0) = 10 \text{ m/s}$.