a.V^2 = X^2 + Y^2 = 0.8^2 + 1.2^2 = 2.08
Vs = 1.442 m/s. = Resultant velocity of the swimmer.
Tan A = 0.8/1.2 = 0.66667
A = 33.7o E. of N.(N33.7E).
d1 = 550m/cos33.7 = 661.1 m[N33.7E] = Dist. to other side.
T1 = d1/Vs = 661.1/1.442 = 458.5 s. To
reach other side.
d2 = 550*Tan A = 550*Tan33.7 = 366.8 m =
Dist. from down-stream to the dock.
T2 = d2/(1.2-0.8) = 366.8/0.4 = 917 s. From down-river to the dock.
T1 + T2=458.5 + 917 = 1376 s.=23 min =
Time to cross the river and swim to the dock.
b. She would land in front of the dock
instead of down-river.
A lifeguard who can swim at 1.2 m/s in still water wants to reach a dock positioned perpendicularly directly across a 550 m wide river.
a) If the current in the river is 0.80 m/s, how long will it take the lifeguard to reach the dock?
b) If instead she had decided to swim in such a way that will allow her to cross the river in a minimum amount of time, where would she land relative to the dock?
3 answers
I am not a student; I am a tutor.
"i aM nOt a sTUDenT i aM a tuToR"
1 answer