A boat can travel 2.60 m/s in still water.If the boat points its prow directly across a stream whose current is 1.10 m/s, what is the velocity (magnitude and direction) of the boat relative to the shore? What will be the position of the boat, relative to its point of origin, after 3.00 s?
a)displacement downstream
b)displacement across the river
velocity relative to shore?
it is: Vbshore=Vriver +Vboat where Vriver is 1.1m/s downstream; Vboat is 2.60across.
To find the velocity of the boat relative to the shore, we can use vector addition.
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.
To find the velocity of the boat relative to the shore, we need to consider the velocity of the boat in still water and the velocity of the current. We can use vector addition to find the resultant velocity.
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.