- #1
EaGlE
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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?
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?
