Question

Skater A passes skater B with a constant velocity of 12 m/s. Three seconds later, skater B starts after skater A with an acceleration of 4 m/s2. How long does it take skater B to catch skater A?

Answer 1

Equations of motion of a skaters are given by

$$x_A=36+12t\\ x_B=2t^2$$

When the skater B catches skater A

$$x_A=x_B\\ 36+12t=2t^2$$

We get

$$t^2-6t-18=0$$ $$t=8.2\:\rm s$$