Q32: Time Interval for Particle C

Particles A and B move towards each other with speed $v = 5 \text{ m/s}$. Initially ($t=0$), the distance $AB = 100 \text{ m}$. The separation between A and B as a function of time is: $$ d_{AB}(t) = |100 – (v_A + v_B)t| = |100 – 10t| $$

Particle C maintains constant distances $AC = 40 \text{ m}$ and $BC = 30 \text{ m}$. For such a triangle ABC to exist, the side lengths must satisfy the Triangle Inequality: $$ |AC – BC| \leq AB \leq AC + BC $$

Substituting the known values: $$ |40 – 30| \leq |100 – 10t| \leq 40 + 30 $$ $$ 10 \leq |100 – 10t| \leq 70 $$

Case 1: Approach Phase ($t < 10$)

Here $100 – 10t > 0$, so $|100 – 10t| = 100 – 10t$. $$ 10 \leq 100 – 10t \leq 70 $$ Subtract 100 from all sides: $$ -90 \leq -10t \leq -30 $$ Divide by -10 (reversing inequalities): $$ 3 \leq t \leq 9 $$

Case 2: Separation Phase ($t > 10$)

Here the particles have crossed each other, so $|100 – 10t| = 10t – 100$. $$ 10 \leq 10t – 100 \leq 70 $$ Add 100 to all sides: $$ 110 \leq 10t \leq 170 $$ Divide by 10: $$ 11 \leq t \leq 17 $$

The condition is satisfied in both intervals.