Question

At $t=0$, one athlete in a race running on a long, straight track with a constant speed $v_1$ is a distance $d_1$ behind a second athlete running with a constant speed $v_2$. (a) Under what circumstances is the first athlete able to overtake the second athlete? (b) Find the time $t$ at which the first athlete overtakes the second athlete, in terms of $d_1,v_1$, and $v_2$. (c) At what minimum distance $d_2$ from the leading athlete must the finish line be located so that the trailing athlete can at least tie for first place? Express $d_2$ in terms of $d_1,v_1$, and $v_2$ by using the result of part (b)

Answer 1

(a) In order for the trailing athlete to be able to catch the leader, his speed ($v_1$) must be greater than that of the leading athlete ($v_2$), and the distance between the leading athlete and the finish line must be great enough to give the trailing athlete sufficient time to make
up the deficient distance, $d$.
(b) During a time interval $t$ the leading athlete will travel a distance $d_2=v_{2}t$ and the trailing athlete will travel a distance $d_1=v_{1}t$. Only when $d_1=d_2+d$ (where $d$ is the initial distance the trailing athlete was behind the leader) will the trailing athlete have caught the leader. Requiring that this condition be satisfied gives the elapsed
time required for the second athlete to overtake the first
$d_1=d_2+d$ or $v_{1}t=v_{2}t+d$
giving
$v_{1}t-v_{2}t=d$ or $t=\frac{d}{(v_1+v_2)}$
(c) In order for the trailing athlete to be able to at least tie for first place, the initial distance $D$ between the leader and the finish line must be greater than or equal to the distance the leader can travel in the time $t$ calculated above (i.e., the time required to overtake the leader). That is, we must require that
$D\ge d_2=v_{2}t=v_2\left[\frac{d}{(v_1-v_2)}\right]$ or $d_2=\frac{v_{2}d}{v_1-v_2}$