Jeffrey wants to swim his girlfriend who is straight across from him, on the other side of a river bank. The river has a current velocity of 4km/h [W]. The width of the river is 300m. If it takes Jeffrey 15 minutes to reach her,
a) how fast should the boy swim to cross the river in 15 minutes?
b) what is his speed with respect to the river, as seen by the girl's father coming after the two with a loaded gun in a boat?
c) What should Jeffrey's heading be in order for him to end up exactly where his girlfriend is?
For question a)I got 4.2 km/h, and for c) I got a heading of 16.7 degrees.
I don't understand how to obtain the answer for b). Please help
To solve question b), we need to find the speed of Jeffrey with respect to the river, as seen by the girl's father.
First, let's calculate the speed of Jeffrey's swim across the river in 15 minutes.
Given:
- Distance across the river (width) = 300m
- Time taken to cross the river = 15 minutes
To calculate the speed, we need to convert the time to hours, as the given velocity of the river is in km/h.
Converting 15 minutes to hours:
15 minutes = 15/60 hours = 0.25 hours
Now, we can calculate Jeffrey's speed across the river using the formula:
Speed = Distance / Time
Speed = 300m / 0.25 hours = 1200m/h
However, we need to convert the speed to km/h for consistency:
1200m/h = 1.2km/h
So, the speed at which Jeffrey should swim across the river to reach his girlfriend in 15 minutes is 1.2km/h.
Now, let's move on to question b):
To calculate Jeffrey's speed with respect to the river, we need to consider the effect of the river's current. Since the current is 4km/h in the west direction (W), we need to subtract this velocity from Jeffrey's swimming velocity.
Jeffrey's net speed with respect to the river = Jeffrey's swimming speed - River's current velocity
Net speed = 1.2km/h - 4km/h
Net speed = -2.8km/h
The negative sign indicates that Jeffrey's speed with respect to the river is in the opposite direction of the river's current.
Therefore, Jeffrey's speed with respect to the river, as seen by the girl's father coming after them, is -2.8km/h.
Note: The negative sign indicates that Jeffrey will appear to be moving backward with respect to the river's current if observed from a stationary point on the riverbank.
I hope this clarifies how to obtain the answer for question b)! If you have any further questions, feel free to ask.
d = 300 m = 0.3km
t = 15 min. = 0.25 h.
a. V = d/t = 0.3km/0.25h = 1.2 km/h
c. Tan A = 4/1.2 = 3.333
A = 73.3o West of North due to wind.
Heading = 73.3o East of North = 16.7o
CCW.
Vw + Vs = 1.2i
-4 + Vs = 1.2i
Vs = 4 + 1.2i = 4.18km/h[16.7o] CCW = 4.18km/h[76.3o] East of North.