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?
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.
"i aM nOt a sTUDenT i aM a tuToR"
I am not a student; I am a tutor.
To determine the answers, we can use the concept of vector addition. Since the lifeguard is swimming across a river, we need to consider the velocity of the current as well.
a) To find the time it takes the lifeguard to reach the dock, we can use the formula:
time = distance / speed
Since the current is perpendicular to the direction the lifeguard wants to swim, it only affects the horizontal component of their velocity. The horizontal component of their velocity is given by:
v_horizontal = v_swim - v_current
where v_swim is the swimming velocity (1.2 m/s) and v_current is the velocity of the current (0.80 m/s).
Using Pythagoras' theorem, we can find the magnitude of the remaining vertical velocity component:
v_vertical = sqrt(v_swim^2 - v_horizontal^2)
Now we can calculate the time it takes to cross the river:
time = distance across river / v_horizontal
Substituting the given values, we have:
time = 550 m / (1.2 m/s - 0.80 m/s)
Now let's do the calculations:
v_horizontal = 1.2 m/s - 0.80 m/s = 0.4 m/s (to the right)
v_vertical = sqrt((1.2 m/s)^2 - (0.4 m/s)^2) = 1.12 m/s (upward)
time = 550 m / 0.4 m/s = 1375 s
So, it will take the lifeguard approximately 1375 seconds
b) To find where the lifeguard would land if she wanted to swim in a way that minimizes the time taken, we need to analyze the vectors involved.
In this case, the lifeguard should swim in a direction that minimizes her displacement (the shortest distance) across the river. This means she should aim to cancel out the effect of the current as much as possible.
Since the current is to the left (opposite to the swimming direction), she should aim slightly upstream to compensate for the current. By doing so, she will experience a net velocity that is more vertically oriented (upstream) and less horizontally oriented (across the river).
Therefore, she will land upstream relative to the dock. The exact position depends on the specifics of the situation and cannot be determined without additional information.