velocity relative to shore?
it is: Vbshore=Vriver +Vboat where Vriver is 1.1m/s downstream; Vboat is 2.60across.
a)displacement downstream
b)displacement across the river
it is: Vbshore=Vriver +Vboat where Vriver is 1.1m/s downstream; Vboat is 2.60across.
Velocity of the boat in still water = 2.60 m/s (directed horizontally)
Velocity of the stream = 1.10 m/s (directed perpendicular to the boat)
To find the resultant velocity, we can use the Pythagorean theorem:
Resultant velocity = sqrt((2.6^2) + (1.1^2)) = sqrt(6.76 + 1.21) = sqrt(7.97) ≈ 2.82 m/s
Now, to find the direction of the resultant velocity, we can use trigonometry:
Theta = arctan(1.10 / 2.60) = arctan(0.423) ≈ 22.9 degrees
Therefore, the velocity of the boat relative to the shore is approximately 2.82 m/s at an angle of 22.9 degrees upstream.
Now, to find the position of the boat relative to its point of origin after 3.00 seconds, we can use the equation:
Displacement = Velocity × Time
a) Displacement downstream:
The boat is moving upstream, so the displacement downstream will be negative:
Displacement downstream = - (2.82 m/s) × (3.00 s) = - 8.46 m
b) Displacement across the river:
Displacement across the river = (1.10 m/s) × (3.00 s) = 3.30 m
Therefore, after 3.00 seconds, the boat will have a displacement of -8.46 m downstream and a displacement of 3.30 m across the river, relative to its point of origin.
The magnitude of the boat's velocity in still water is given as 2.60 m/s. The magnitude of the current's velocity is given as 1.10 m/s.
To find the magnitude of the boat's velocity relative to the shore, we use the Pythagorean theorem:
v_relative = sqrt(v_boat^2 + v_current^2)
v_relative = sqrt((2.6 m/s)^2 + (1.1 m/s)^2)
v_relative = sqrt(6.76 m^2/s^2 + 1.21 m^2/s^2)
v_relative = sqrt(7.97 m^2/s^2)
v_relative ≈ 2.82 m/s
So, the magnitude of the boat's velocity relative to the shore is approximately 2.82 m/s.
To find the direction of the boat's velocity relative to the shore, we use trigonometry. The angle between the boat's velocity and the current's velocity is given by:
θ = atan(v_boat / v_current)
θ = atan(2.60 m/s / 1.10 m/s)
θ ≈ 67.42°
So, the boat's velocity is directed at an angle of approximately 67.42° with respect to the current.
Now, let's find the displacements downstream and across the river after 3.00 s.
Downstream displacement:
s_downstream = v_relative * t
s_downstream = 2.82 m/s * 3.00 s
s_downstream ≈ 8.46 m
The downstream displacement of the boat after 3.00 s is approximately 8.46 m.
Across the river displacement:
s_across = v_current * t
s_across = 1.10 m/s * 3.00 s
s_across = 3.30 m
The across-the-river displacement of the boat after 3.00 s is 3.30 m.
So, the answers to the questions are:
a) The displacement downstream is approximately 8.46 m.
b) The displacement across the river is 3.30 m.