Question 468784: Hello I have a uniform motion problem involving rational expressions, I normally get these but, I can't find the set up for this one. I know the whole d=rt, r=d/t and t=r/d but I can't seam to figure out whats what,anyways here it goes:
"Suppose that Jeremy Shockey of the New Orlean Saints can run 100yards in 12 seconds. Further suppose Brian Urlacher of the Chicago Bears can run 100yards in 9 seconds. Suppose that Shockey catches a pass at his own 20-yard line in stride and starts running away from Urlacher, who is at the 15-yard line directly behind Shockey. At what yard line will Urlacher catch up to Shockey and tackle him?"
Thanks for the help!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! THERE IS A LONG WAY TO CALCULATE THIS AND THERE IS A SHORT WAY TO CALCULATE THIS.
FIRST THE LONG WAY.
shockey rate of travel is 100 yards in 12 seconds = 100/12.
urlacher rate of travel is 100 yards in 9 seconds = 100/9.
urlacher will catch shockey in 1.8 seconds.
in 1.8 seconds, shockey will have traveled 1.8 * (100/12) = 15 yards.
in 1.8 seconds, urlacher will have traveled 1.8 * (100/9) = 20 yards.
the formulas that we used to solve this problem are:
R1*T = D for shockey
R2*T = D+5 for urlacher.
We know R1 and R2, but we don't know D or T, although we do know that urlacher had to travel 5 more yards than shockey in order to catch him.
Since T will be the same for both, we solve each of these equations in terms of T to get:
T = D/R1
T = (D+5)/R2
since the expressions on the right hand side of each equation are both equal to T, they we can set these expressions equal to each other to get:
D/R1 = (D+5)/R2
Cross multiply to get:
D*R2 = (D+5)*R1
Expand the expression on the right side of the equation to get:
D*R2 = D*R1 + 5*R1
subtract D*R1 from both sides of the equation to get:
D*R2 - D*R1 = 5*R1
factor out the D from the left side of the equation to get:
D*(R2-R1) = 5*R1
divide both sides of the equation by (R2-R1) to get:
D = (5*R1)/(R2-R1)
This formula will get us D from which we can then derive T.
Since R1 = (100/12) and R2 = (100/9), this formula becomes:
D = (5*(100/12)) / ((100/9) - (100/12)) which gets us:
D = 15
This makes D + 5 = 20
shockey runs 15 yards in the same time that urlacher runs 20.
since we know D, we can solve for T.
Either equation should do it because the time is the same in both.
We'll use Shockey's equation.
That is R1 * T = D which becomes:
(100/12) * T = 15
Divide both sides of this equation by (100/12) to get:
T = 15/(100/12) which becomes:
T = 1.8 seconds.
shockey runs 15 yards in 1.8 seconds.
urlacher runs 20 yards in 1.8 seconds.
urlacher catches shockey in 1.8 seconds.
***********************************************
NOW THE SHORT WAY
***********************************************
We already derived our formulas above using rate * time = distance.
They are:
R1 * T = D for shockey.
R2 * T = D + 5 for urlacher.
we want to solve these equations simultaneously since the same solution will apply to both equations.
we subtract the first equation from the second equation to get:
R2*T - R1*T = 5
We factor out the T to get:
T * (R2-R1) = 5
We divide both sides of this equation by (R2-R1) to get:
T = 5/(R2-R1)
Since R2 = 100/12 and R1 = 100/9, we substitute in this equation to get:
T = 5/((100/12)-(100.9)) to get:
T = 1.8
Notice that we solved for T but did not solve for D.
If we want D, we still have to go back in to either equation to solve for D.
If we used the second equation (the first would have been a little easier), we would get:
R2 * T = D + 5
We know R2 and we know T so we substitute to get:
(100/9) * 1.8 = D + 5
we subtract 5 from both sides of this equation to get:
((100/9)*1.8) - 5 = D
We solve for D to get:
D = 15
THIS PROBLEM CAN BE SOLVED EITHER WAY AND YOU'LL GET THE SAME ANSWER.
THE SHORT WAY SEEMS EASIER IF YOU RECOGNIZE THAT YOU CAN DO IT THAT WAY.
|
|
|
