meg rowed her boat upstream a distance of 5 miles

Question

meg rowed her boat upstream a distance of 5 miles then rowed back to the starting point. the total time of the trip was 6 hours. if the rate of the current was 2mph, find the average speed of the boat relative to the water.

Answer 1

Meg, Mathematically the information given is:

Tu + Td = 6 hours
D = 5 miles
Current Speed is 2 mph ( assumed downstream of course!)
Boat Speed = S = ???
Avg Speed = Sa = ???

Su = S – 2 upstream speed diminished by current (0)
Sd = S + 2 downstream speed enhanced by current

Su*Tu = 5 miles Tu is the upstream travel time (1)
Sd*Td = 5 miles Td is the downstream travel time (2)

Isolate Tu from (0) & (1):
(S-2)*Tu =5
Tu = 5 / (S-2)

Use Tu + Td = 6 with the previous result:
5/(S-2) + Td = 6
Isolate Td:
Td = (6*S-17)/(S-2)

Substitute this Td expression back into (2), (recall that Sd=S+2):
Sd*Td = 5
(S + 2)*(6*S-17)/(S-2)=5

Simplify into a quadratic expression and factor:

6*S^2 -10*S -24 = 0

( 3*S + 4) * (2S – 6) =0 (Check: 6*S^2 + 8*S - 18*S – 24 = 0 OK!)
Solutions to the quadratic equation:
(3*S + 4) = 0 or S = -4/3 (an unusual speed!)
(2*S – 6) = 0 or S = 3
Check the answers:

S=3
Su = S-2 = 3-2 = 1 mph (3)
Sd = S+2= 3+2= 5 mph (4)

Tu = 5/1 = 5 hrs
Td = 5/5 = 1 hrs
Tot time 6 hrs OK!

S= -4/3
Su=S-2=(-4/3) -2=-10/3
Sd=S+2= (-4/3) +2 = -4/3+6/3 =2/3

Tu = 5/(-10/3)= -15/10
Td = 5/(2/3) = +75/10
Tot time (75-15)/10 = 60/10 + 6 hrs OK but STRANGE!

Caveat: The negative speeds are mathematically correct and lead to the correct total travel time. But note that the Tu=-10/15 is a negative time which is not a real answer. So the correct answers are (3) & (4) and the correct Avg Speed = Sa = 1 + 5 = 6 mph.
BruceS

Answer 2

so she went from point A to point B and back again to point A (kinda pointless, huh) let's say v is the speed of the boat relative to the water, then downstream speed = v + 2 upstream speed = v - 2 you need this equation (v = d / t) (speed = distance over time) we know the total time is 6 hours, so we just need to add the 2 leg times up, time from a to b, and time from b to a t1 + t2 = total time 5/(v-2) + 5/(v+2) = 6 solve for v... 5(v+2) + 5(v-2) = 6(v+2)(v-2) 5v+10 + 5v-10 = 6(v^2-4) 10v = 6v^2 - 24...bring everything to one side 0 = 6v^2 - 10v - 24 --> 3v^2 - 5v - 12 = 0 using the quadratic formular, we get v = -4/3 or v = 3 we want the speed that makes sense, so we choose the positive one v = 3