# college physics

One athlete in a race runnng on a long, straight track with a constanct speed v1 is a distance d behind a second athlete running with a constant speed v2.

(a) Under what circumstances is the first athlete able to overtake the second athlete? (Use v_1 for v1, v_2 for v2, and d as appropriate.)

(b) Find the time t it takes the first athlete to overtake the second athlete, in terms of d, v1, and v2.

(c) At what minimum distance d2 from the leading athlete must the finish line be located so that the trailing athlete can at least tie for the first place? Express d2 in terms of d, v1, and v2 by using the result of part (b).

1. a.
Of course, one way for athlete 1 to overtake athlete 2 is for him to accelerate (that is, if v2 > v1).
But since the given are constant speeds, surely v2 MUST BE LESS THAN v1 in order for athlete 1 to overtake athlete 2, WITHOUT accelerating.

b.
Recall that the distance travelled by an object is just speed x time:
d = vt
Athlete 2 is ahead by a distance, d, and after some time, t, their distance travelled will be equal (that is, athlete 1 overtakes athlete 2):
d + (v2)(t) = (v1)(t)
Solving for t,
t = d/(v1 - v2)

c.
Now there is an additional condition. Athlete 2 is d2 away from the finish line, while athlete 1 is (d + d2) away from the finish line.
d2 = (v2)t
from b., we substitute the expression for t:
d2 = (v2)*(d)/ (v1 - v2)

Hope this helps~ :3

posted by Jai

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