Answer to Question #239552 in Electrical Engineering for may

Question

Two runners are competing in a 400-meter dash. The first run at 6 m/s and maintains this speed. The second runner starts at 10 m/s and decelerates at 1/8 m/s². Who will finish first? What will be their time difference?

Answer 1

The time required for the first competitor to finish is

$$t_1=\frac D{v_1}=66.67\text{ s}.$$

The time required for the second runner is

$$D=v_2t_2-\frac{at^2_2}2,\\\space\\ t_2^2\frac a2-t_2v_2+D=0,\\\space\\ t_2=\frac{v_2\pm\sqrt{v_2^2-2aD}}{a}=80\text{ s}.\\\space\\ \Delta t=t_2-t_1=13.33\text{ s}.$$

The second runner will be 13.33 s behind the winner.