Answer:
a)40 meters
b)2 m/s
c)2 m/s
d)0 m/s
e)45 degrees northeast
Step-by-step explanation:
a) The displacement from C to A is the distance directly across the river, which is 40 meters.
b) The speed of the boat as seen by people standing at A is the magnitude of the boat's velocity vector, which is equal to the displacement divided by the time taken:
Speed = displacement / time = 40 m / 20 s = 2 m/s.
c) Let v be the speed of the water in the river. The boat is moving at right angles to the flow of the river, so the water exerts a perpendicular force on the boat. The time taken for the boat to travel from A to C is 20 seconds, during which time the boat will have been carried downstream by the river by a distance equal to v times the time taken.
Distance carried downstream = v × time = v × 20 m.
Since the boat landed at C, which is directly across the river from A, the distance it traveled horizontally is 40 meters. Therefore:
40 m = (boat speed) × (time taken) = (boat speed) × 20 s.
Hence, the speed of the boat is:
Boat speed = 40 m / 20 s = 2 m/s.
So, we have two equations:
Distance carried downstream = v × 20 m
Boat speed = 2 m/s
From the first equation, we get:
v × 20 m = 40 m
Therefore, the speed of the water in the river is:
v = 40 m / 20 m = 2 m/s.
d) The speed of the boat as seen by a fish drifting with the river is the difference between the speed of the boat and the speed of the water in the river:
Boat speed - Water speed = 2 m/s - 2 m/s = 0 m/s.
So, the speed of the boat as seen by a fish drifting with the river is zero.
e) The boat should head in a direction that makes its velocity vector point directly from A to B. Since A and B are directly opposite each other, this means the velocity vector should be perpendicular to the line connecting A and B.
We know the boat's velocity vector has a magnitude of 2 m/s and is at right angles to the velocity vector of the water in the river, which has a magnitude of 2 m/s. So, we can draw a vector diagram with the velocity vector of the boat pointing straight up and the velocity vector of the water pointing straight to the right. The vector connecting the tail of the water velocity vector to the head of the boat velocity vector will then point directly from A to B.
The angle between the boat's velocity vector and the line connecting A and B can be found using trigonometry. Let θ be this angle. Then:
tan(θ) = (boat speed) / (water speed) = 2 m/s / 2 m/s = 1.
Taking the inverse tangent of both sides gives:
θ = tan^(-1)(1) = 45°.
So, the boat should head in a direction 45 degrees to the right of straight up, or northeast.