Respuesta :
rate of tight end = 100 yards in 10 seconds = 100/10.
rate of defensive back = 100 yards in 12 seconds = 100/12.
tight end starts at 20 yards.
defensive back starts at 15 yards.
defensive back needs to catch tight end at the Z yard line hopefully before the tight end makes a touchdown. We assume the distance is a Z yard line
defensive back runs 100/10 yards per second from 15 yard line to get to the Z yard line.
tight end runs 100/12 yards per second from 20 yard line to get to the Z yard line.
let s = number of seconds to be run which needs to be the same for both players.
formula for defensive back is 15 + (100/10)*s = Z
formula for tight end is 20 + (100/12)*s = Z
since they both = Z, then they are equal to each other.
we get
15 + (100/10)*s = 20 + (100/12)*s
subtracting 15 from both sides of the equation yields
(100/10)*s = 20 - 15 + (100/12)*s
subtracting (100/12)*s from both sides of the equation yields
(100/10)*s - (100/12)*s = 20 - 15.
solving we get
10*s - 8.3333333*s = 5 which becomes
1.1666666667*s = 5 which becomes
s= 5/1.666666667 = 3.
they will both reach the Z yard line in 3 seconds.
tight end will travel (100/12)*3 = 25 yards
defensive back will travel (100/10)*3 = 30 yards.
tight end starts at the 20 and travels 25 yards to get to the 45 yard line.
defensive back starts at the 15 and travels 30 yards to get to the 45 yard line in the same amount of time.
defensive back will catch the tight end at the 45 yard line.
