How much time does it take the walker to make the round trip?

[EaGlE]

Two piers, A and B , are located on a river: B is 1500 m downstream from A . Two friends must make round trips from pier A to pier B and return. One rows a boat at a constant speed of 4.00 km/h relative to the water; the other walks on the shore at a constant speed of 4.00 km/h. The velocity of the river is 2.80 km/h in the direction from A to B .

what does "relative to the water" mean?

1.) How much time does it take the walker to make the round trip?
answer must be in mins

2.) How much time does it take the rower to make the round trip? answer must be in mins

my work:

Given:
x=1500m
v(b) = 4.00 km/h
v(w) = 4.00 km/h
v(r) = 2.80

t1 = (1500)/(4+2.80) = 220.588secs <--- downstream time
t2 = (1500)/(4-2.80) = 1250 secs <--- upstream time

t(t) = 1470.588s <--- total time

how would i solve #1 ?

im thinking that i would need the distance formula

x(t) = x(0) + v(0)t + 1/2at^2
2(1500) = 0 + 2.80t + 0 (x(0) = 0, because at t=0, x=0. and a=0 because of constant speed)

3600= 2.80t... nevermind, it doesn't look right, can someone help?

 

[Doc Al]

2.) How much time does it take the rower to make the round trip? answer must be in mins

my work:

Given:
x=1500m
v(b) = 4.00 km/h
v(w) = 4.00 km/h
v(r) = 2.80

t1 = (1500)/(4+2.80) = 220.588secs <--- downstream time
t2 = (1500)/(4-2.80) = 1250 secs <--- upstream time

t(t) = 1470.588s <--- total time

Basic idea is OK, but you are mixing up units. It's easier than you think. I'll do the first part, the time from A to B:
t1 = D/V = (1.5 Km)/(6.28 Km/hour) = 0.24 hours
How many minutes is that? You take it from here.

how would i solve #1 ?

Exactly the same way, only now the speed is just 4.0 Km/hour both ways. So, t = D/V ...

 

[HallsofIvy]

"relative to the water" means that that is his speed treating the water as it were not flowing. The actual or "true" speed is the water's speed added to his boat's speed when he is going down stream with the current, and subtracted from his boat's speed when he is going upstream against the current.

 

[EaGlE]

thank you, works perfectly