Question

a sprinter who averaged 30 ft/sec completed a race 1 sec before another sprinter who averaged 28 ft/sec. what was the distance of the race?

Answer 1

420 ft

Answer 2

x = distance travelled by any of the racer
distance = speed x time
x = 30 ft/sec (t1)
x = 28ft/sec (t2)
t1 = t2-1
30ft/sec (t2-1) = 28ft/sec(t2)
30t2 - 30 = 28t2
30t2-28t2=30
2t2 = 30
t2 = 15 secs
x = 28(15) = 420 ft