Question 158215This question is from textbook Sullivan-College Algebra
: #37 Football: A tight end can run the 100-yd dash in 12 seconds. A defensive back can do it in 10 seconds . The T.E. catches a pass at his own 20-yd line with the D.B. at the 15 yd line . If no other players are nearby, at what yd line will the defensive back catch up to the tight end? (Hint: at t=0, the D.B. is 5 yds behind the T.E.) Answer in back of book is: DB catches up to tight end at the TE's 45 yd line.
#49 A bathroom tub will fill in 15 minutes with both faucets open and the stopper in place. With both faucets closed and the stopper removed, the tub will empty in 20 minutes. How long will it take for the tub to fill if both faucets are open and the stopper is removed. Answer in the back of the book is : the tub will fill in one hour.
Thank you,
Shelia
This question is from textbook Sullivan-College Algebra
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! #37:
rate of tight end = 100 yards in 10 seconds = 100/10.
rate of defensive back = 100 years 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 P yard line hopefully before the tight end makes a touchdown.
defensive back runs 100/10 yards per second from 15 yard line to get to the P yard line.
tight end runs 100/12 yards per second from 20 yard line to get to the P yard line.
let x = number of seconds to be run which needs to be the same for both players.
formula for defensive back is 15 + (100/10)*x = P
formula for tight end is 20 + (100/12)*x = P
since they both = P, then they are equal to each other.
we get
15 + (100/10)*x = 20 + (100/12)*x
subtracting 15 from both sides of the equation yields
(100/10)*x = 20 - 15 + (100/12)*x
subtracting (100/12)*x from both sides of the equation yields
(100/10)*x - (100/12)*x = 20 - 15.
solving we get
10*x - 8.3333333*x = 5 which becomes
1.1666666667*x = 5 which becomes
x = 5/1.666666667 = 3.
they will both reach the P 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.
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#49:
tub will fill in 15 minutes when both faucets are open and stopper is in.
tub will drain in 20 minutes when both stopper is out.
how long does it take the tub to fill if both faucets are open and stopper is out?
let G = number of gallons that the tub holds.
let R(f) = rate of filling the tub.
let R(e) = rate of emptying the tub.
formulas appear to be
R(f)*15 = G when stopper is in and both faucets are running.
R(e)*20 = G when stopper is out and both faucets are closed.
when both faucets are open and the the stopper is out, the rate of filling the tub is working against the rate of emptying of the tub, so
(R(f) - R(e))*T = G
where T is the amount of time required.
in order to solve this equation we need to remove one of the unknowns.
if R(f)*15 = G and R(e)*20) = G, then
R(f)*15 = R(e)*20 which becomes
R(f) = (20/15)*R(e)
using that equality, the equation
(R(f) - R(e))*T = G becomes
((20/15)*R(e) - R(e))*T = G
this becomes
(1/3)*R(e)*T = G
we still have 2 unknowns, so we use another equality to remove the second unknown.
the second unknown is G.
we have the equation R(e)*20 = G to draw from.
that changes our equation to
(1/3)*R(e)*T = R(e)*20
now we have one unknown and can solve the equation.
it becomes
((1/3)*R(e)*T)/R(e) = 20
the R(e) cancels out (same value in numerator and denominator) so the equation becomes
(1/3)*T=20
multiplying both sides of the equation by 3 we get
T = 60.
since T is in minutes, the answer is 60 minutes.
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