V = 1.2 - 0.80i = 1.44m/s[-33.7o] = 1.44m/s[33.7o] S. of E. = 326.3o CCW.
a. Cos326.3 = 550/d.
d1 = 550/Cos326.3 = 661 m. across due to current.
V*T1 = 661.
1.44 * T1 = 661,
T1 = 459 s. = 7.7 min. to cross river.
Tan326.3 = d2/550.
d2 = 550*Tan326.3 = 367 m. Downstream from dock.
d2 = V2 * T2 = 367.
(1.2-0.8) * T2 = 367,
T2 = 917 s. = 15.3 min. from downstream to dock.
T1 + T2 = 7.7 + 15.3 = 23 min. = Total time to reach dock.
b. Directly in front of the dock.
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.
If the current in the river is 0.80 m/s, how long will it take the lifeguard to reach the dock?
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?
1 answer
1 answer