SOLUTION: 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 1

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(52793)   (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): You can put this solution on YOUR website!

The response from the other tutor shows a formal algebraic solution using a variation of what is probably the most common method, and very likely the method taught in most references.

I prefer a different solution method for a problem like this, like the following.

The tight end and the defensive back can run the 100-yard dash is 12 seconds and 10 seconds, respectively. Since those distances are the same, the ratio of their speeds is 10:12 = 5:6.

When the tight end catches the pass, he is 5 yards past the defensive back. So he will catch up to the tight end when he has run 5 yards farther than the tight end.

Let x = # yards the tight end runs after catching the pass
Then x+5 = # of yards the defensive back runs

Since the times are the same, the ratio of numbers of yards is the same as the ratio of their speeds:

From the time the tight end catches the pass until the defensive back catches up to him, the tight end runs x = 25 yards and the defensive back runs x+5 = 30 yards. Since the tight end catches the pass at his own 20-yard line, the yard line where the defensive back catches him is 20+25 = 45.

ANSWER: at the tight end's own 45-yard line

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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