A canoe has a velocity of 0.490 m/s, 45 degrees southeast relative to the earth. The canoe is on a river that is flowing at 0.600 m/s east relative to the earth.
Find the magnitude of the velocity vector v of the canoe relative to the river.
Express your answer in meters per second.
To find the magnitude of the velocity vector v of the canoe relative to the river, we can use the Pythagorean theorem.
The velocity of the canoe relative to the earth is given as 0.490 m/s at a direction 45 degrees southeast. In terms of components, the magnitude of the velocity vector of the canoe relative to the earth can be represented as:
Vc_e = 0.490 m/s
Vc_e_x = Vc_e * cosθ = 0.490 * cos(45°) = 0.490 * √2/2 = 0.345 m/s
Vc_e_y = Vc_e * sinθ = 0.490 * sin(45°) = 0.490 * √2/2 = 0.345 m/s
The velocity of the river relative to the earth is given as 0.600 m/s east. In terms of components, the magnitude of the velocity vector of the river relative to the earth can be represented as:
Vr_e = 0.600 m/s
Vr_e_x = Vr_e = 0.600 m/s
Vr_e_y = 0
The velocity vector of the canoe relative to the river can be represented as:
Vc_r_x = Vc_e_x - Vr_e_x = 0.345 m/s - 0.600 m/s = -0.255 m/s
Vc_r_y = Vc_e_y - Vr_e_y = 0.345 m/s - 0 = 0.345 m/s
To find the magnitude of the velocity vector of the canoe relative to the river (Vc_r), we can use the Pythagorean theorem:
Vc_r = sqrt(Vc_r_x^2 + Vc_r_y^2)
Vc_r = sqrt((-0.255 m/s)^2 + (0.345 m/s)^2)
Vc_r = sqrt(0.065025 + 0.119025)
Vc_r = sqrt(0.18405)
Vc_r ≈ 0.429 m/s
Therefore, the magnitude of the velocity vector v of the canoe relative to the river is approximately 0.429 m/s.
To find the magnitude of the velocity vector v of the canoe relative to the river, we can apply the concept of vector addition.
First, let's represent the velocity of the canoe relative to the earth as vector C and the velocity of the river relative to the earth as vector R.
Given:
Velocity of the canoe relative to the earth (C) = 0.490 m/s, 45 degrees southeast
Velocity of the river relative to the earth (R) = 0.600 m/s, east
To find the velocity of the canoe relative to the river, we need to subtract the velocity of the river from the velocity of the canoe: v = C - R.
Step 1: Convert the velocity vector of the canoe from polar form to rectangular form.
Given:
Magnitude of the velocity of the canoe (C_mag) = 0.490 m/s
Angle of the velocity of the canoe (C_angle) = 45 degrees southeast
To convert the vector from polar form to rectangular form, we can use the following equations:
C_x = C_mag * cos(C_angle)
C_y = C_mag * sin(C_angle)
Step 2: Convert the velocity vector of the river from polar form to rectangular form.
Given:
Magnitude of the velocity of the river (R_mag) = 0.600 m/s
Angle of the velocity of the river (R_angle) = 0 degrees (east)
To convert the vector from polar form to rectangular form, we can use the following equations:
R_x = R_mag * cos(R_angle)
R_y = R_mag * sin(R_angle)
Step 3: Subtract the rectangular components of the velocity of the river from the rectangular components of the velocity of the canoe:
v_x = C_x - R_x
v_y = C_y - R_y
Step 4: Calculate the magnitude of the velocity vector v:
v_mag = sqrt(v_x^2 + v_y^2), where the sqrt function represents the square root.
Now let's calculate:
C_x = (0.490 m/s) * cos(45 degrees) = 0.490 * 0.7071 = 0.346 m/s
C_y = (0.490 m/s) * sin(45 degrees) = 0.490 * 0.7071 = 0.346 m/s
R_x = (0.600 m/s) * cos(0 degrees) = 0.600 * 1 = 0.600 m/s
R_y = (0.600 m/s) * sin(0 degrees) = 0.600 * 0 = 0 m/s
v_x = 0.346 m/s - 0.600 m/s = - 0.254 m/s
v_y = 0.346 m/s - 0 m/s = 0.346 m/s
v_mag = sqrt((-0.254 m/s)^2 + (0.346 m/s)^2) = sqrt(0.064516 + 0.119316) = sqrt(0.183832) = 0.428 m/s
Therefore, the magnitude of the velocity vector v of the canoe relative to the river is 0.428 m/s.
Draw the triangle of vectors, ABC, where
AB represents the river velocity, 0.6 m/s due east, BC is the canoe velocity, 0.49 m/s due south-east.
You are looking for the resultant vector AC.
You can solve this by the law of cosines, which is
AC²=AB²+BC²-2AB.BC.cos(135°)
or by summing components along the x- and y-axes (East and north).
Resultant:
x-component=0.6+0.49cos(-45°) m/s
y-component=0.49sin(-45°) m/s
tan(θ)=y-component/x-component