Question 1208986: A tight end can run the 100-yard dash in 12 seconds. A defensive back can do it in 10 seconds. The tight end catches a pass at his own 20-yard line with the defensive back at the 15-yard line. If no other players are nearby, at what yard line will the defensive back catch up to the tight end?
Hint: At time t = 0, the defensive back is 5 yards behind the tight end.
Found 4 solutions by ikleyn, MathTherapy, greenestamps, math_tutor2020:
Answer by ikleyn(52790)   (Show Source):
You can put this solution on YOUR website! .
Is it for football or basketball ? Or for table tennis ?
Or, may be, for swimming ?
Which guidebook should I study to understand the meaning of this post ?
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Normally, Math composers create Math problems to teach students on Math.
Quasi-"math composers" do create problems to demonstrate how smart they are.
The distance between these two categories is an abyss.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
A tight end can run the 100-yard dash in 12 seconds. A defensive back can do it in 10 seconds. The tight end catches a pass at his own
20-yard line with the defensive back at the 15-yard line. If no other players are nearby, at what yard line will the defensive back
catch up to the tight end?
Hint: At time t = 0, the defensive back is 5 yards behind the tight end.
The tight end can run the 100-yard dash in 12 seconds, so his speed = 
The defensive back can run the 100-yd dash in 10 seconds, so his speed = 
The tight end catches the pass at HIS OWN 20-yard line with the defensive back at the 15-yard line, and so, the
tight end is 5 yards ahead of the defensive back.
Let the distance that the tight end (SLOWER player) needs to travel to get to the catch-up point, be D
Since the defensive back (FASTER player) is 5 yards BEHIND the tight end, distance defensive back needs to travel to get
to the catch-up POINT/catch up to the tight end, is (D + 5) yards
Since BOTH will get to the catch-up point at the same time, we EQUATE their respective times to get
the following TIME equation: 
2(3D) = 5(D + 5) ----- Cross-multiplying
6D = 5D + 25
6D - 5D = 25
Distance tight end needs to travel to get to catch-up point, or D = 25 yards
Location where tight end caught the ball: HIS OWN 20-yard line
Distance tight end travels to get to catch-up point: 25 yards
The defensive back caught up to the tight end at the (20 + 25)-MARK, which is the tight end's OWN 45 YARD-LINE.
Answer by greenestamps(13200)  (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answer: tight end's own 45 yard line
Explanation:
TE = tight end
DB = defensive back
Let's find the TE's speed
distance = rate*time
d = r*t
r = d/t
r = (100 yards)/(12 seconds)
r = (100/12) yards per second
r = (25/3) yards per second
r = 8.3333 yards per second approximately
DB's speed
r = d/t
r = (100 yards)/(10 seconds)
r = (100/10) yards per second
r = 10 yards per second
After x number of seconds, the TE travels a distance of (25/3)x yards.
Meanwhile the DB travels 10x yards.
The gap between the players is initially 5 yards.
Whatever distance the TE travels, the DB must travel that distance plus 5 extra yards to catch up.
For example, if the TE travels 10 yards, then the DB must travel 10+5 = 15 yards to catch up to the TE.
DB's travel distance = (TE's travel distance) + 5 yards
10x = (25/3)x+5
3*10x = 3*( (25/3)x+5 )
30x = 25x+15
30x-25x = 15
5x = 15
x = 15/5
x = 3
The DB catches the TE at exactly the x = 3 second timestamp.
The DB catches (and perhaps tackles) the TE exactly 3 seconds after the TE caught the ball.
After 3 seconds elapse, the TE traveled (25/3)*x = (25/3)*3 = 25 yards.
The TE goes from the 20 yard line to the 20+25 = 45 yard line.
The TE has not crossed the midway marker since 45 < 50.
Therefore the TE is still on his side of the field.
Meanwhile the DB runs 10*x = 10*3 = 30 yards
The DB goes from the 15 yard line to the 15+30 = 45 yard line
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