SOLUTION: A boat traveled downstream a distance of 50 mi and then came right back. If the speed of the current was 20 mph and the total trip took 3 hours and 20 minutes, find the average spe

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: A boat traveled downstream a distance of 50 mi and then came right back. If the speed of the current was 20 mph and the total trip took 3 hours and 20 minutes, find the average spe      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1066350: A boat traveled downstream a distance of 50 mi and then came right back. If the speed of the current was 20 mph and the total trip took 3 hours and 20 minutes, find the average speed of the boat relative to the water.
Found 4 solutions by josgarithmetic, Boreal, josmiceli, MathTherapy:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
               SPEED        TIME           DISTANCE

DOWN           r+20         50%2F%28r%2B20%29             50

UP             r-20         50%2F%28r-20%29             50

TOTAL                     3%261%2F3


Solve this for r.
50%2F%28r%2B20%29%2B50%2F%28r-20%29=3%261%2F3

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
speed of boat is x
speed of current is 20 mph
x+20* t1=50, so t1=50/(x+20)
x-20*t2=50, so t2=50/(x-20)
50/(x+20)+50/(x-20)=10/3, where 10/3 is hours.
multiply everything by (x+20)(x-20 or x^2-400
50(x-20)+50(x+20)=(10/3)x^2-4000/3; the left side is equal to 100x
(10/3)x^2-100x-4000/3=0
multiply everything by 3 then divide everything by 10.
10x^2-300x-4000=0
x^2-30x-400=0
(x-40)(x+10)=0
x=40 mph, only positive root ANSWER.
goes 60 mph downstream and takes 5/6 hr or 50 minutes
goes 20 mph upstream and takes 2.5 hours or 150 minutes.
That takes 200 minutes, which is 3h20m

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = the speed of the boat with no current in mi/hr
+s+%2B+20+ = the speed of the boat going downstream in mi/hr
+s+-+20+ = the speed of the boat going upstream in miles/hr
Let +t+ = time in hrs to go downstream
+10%2F3+-+t+ = time in hrs to go upstream
-----------------------------------------
Going downstream:
(1) +50+=+%28+s+%2B+20+%29%2At+
Going upstream:
(2) +50+=+%28+s+-+20+%29%2A%28+10%2F3+-+t+%29+
-----------------------------------
(2) +50+=+%2810%2F3%29%2As+-+200%2F3+-+s%2At+%2B+20t+
(2) +150+=+10s+-+200+-+3s%2At+%2B+60t+
and
(1) +t+=+50%2F%28+s%2B20+%29+
Plug (1) into (2)
(2) +150+=+10s+-+200+-+3s%2At+%2B+60t+
(2) +150+=+10s+-+200+%2B+t%2A%28+60+-+3s+%29+
(2) +150+=+10s+-+200+%2B+%28+50%2F%28+s%2B20+%29%29%2A%28+60+-+3s+%29+
(2) +350+=+10s+%2B+%28+50%2F%28+s%2B20+%29%29%2A%28+60+-+3s+%29+
(2) +350%2A%28+s%2B20+%29+=+10s%2A%28+s%2B20+%29+%2B+50%2A%28+60+-+3s+%29+
(2) +350s+%2B+7000+=+10s%5E2+%2B+200s+%2B+3000+-+150s+
(2) +10s%5E2+-+300s+-+4000+=+0+
(2) +s%5E2+-+30s+-+400+=+0+
+s+=+%28+-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+1+
+b+=+-30+
+c+=+-400+
+s+=+%28+-%28-30%29+%2B-+sqrt%28+%28-30%29%5E2-4%2A1%2A%28-400%29+%29%29%2F%282%2A1%29+
+s+=+%28+30+%2B-+sqrt%28+900+%2B+1600+%29%29%2F2+
+s+=+%28+30+%2B-+sqrt%28+2500+%29+%29%2F2+
+s+=+%28+30+%2B+50+%29+%2F+2+
+s+=+80%2F2+
+s+=+40+ ( can't use the negative solution )
--------------------
The speed of the boat with no current is 40 mi/hr
--------------------
check:
(1) +50+=+%28+s+%2B+20+%29%2At+
(1) +50+=+%28+40+%2B+20+%29%2At+
(1) +50+=+60t+
(1) +t+=+5%2F6+
and
(2) +50+=+%28+s+-+20+%29%2A%28+10%2F3+-+t+%29+
(2) +50+=+%28+40+-+20+%29%2A%28+10%2F3+-+t+%29+
(2) +50+=+20%2A%28+10%2F3+-+t+%29+
(2) +50+=+200%2F3+-+20t+
(2) +150+=+200+-+60t+
(2) +60t+=+50+
(2) +t+=+5%2F6+
OK




Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!

A boat traveled downstream a distance of 50 mi and then came right back. If the speed of the current was 20 mph and the total trip took 3 hours and 20 minutes, find the average speed of the boat relative to the water.
Let speed in still water be S

Then average speed downstream = S + 20

Average speed upstream: S – 20

We then get the following TIME equation: 50%2F%28S+%2B+20%29+%2B+50%2F%28S+-+20%29+=+3%2620%2F60

50%2F%28S+%2B+20%29+%2B+50%2F%28S+-+20%29+=+10%2F3

50(3)(S – 20) + 50(3)(S + 20) = 10(S + 20)(S – 20) ------- Multiplying by LCD, 3(S  + 20)(S – 20)

150S+-+%223%2C000%22+%2B+150S+%2B+%223%2C000%22+=+10%28S%5E2+-+400%29

300S+=+10%28S%5E2+-+400%29

10%2830S%29+=+10%28S%5E2+-+400%29

30S+=+S%5E2+-+400

S%5E2+-+30S+-+400+=+0

(S - 40)(S + 10) = 0

S, or speed of boat in still water = highlight_green%28matrix%281%2C2%2C+40%2C+mph%29%29   	OR   	  S = - 10 (ignore)