A 200-m-wide river flows due east at a uniform speed of 4.9 m/s. A boat with a speed of 9.0 m/s relative to the water leaves the south bank pointed in a direction 33o west of north. What is the (a) magnitude and (b) direction of the boat's velocity relative to the ground? Give the direction as the angle of the velocity from due north, positive if to the east and negative if to the west. (c) How long does it take for the boat to cross the river?
(a) Add the water velocity vector to the boat velocity vector with respect to the water. Get the magnitude of the resultant, sqrt (Vx^2 + Vy^2)
Vx is the downstream component (east), Vy is the cross-stream component (north)
(b) Get the direction of the resultant. It is the arctangent of the
Vx/Vy ratio
(c) Time to cross = (River width)/Vy
a,b. 33o W. of N.=123o CCW from +x-axis.
Vr = 9[123o] + 4.9 = -4.9+7.55i + 4.9 = 7.55i = 7.55m/s[90o] = Due North.
c. V*t = 200m, 7.55t = 200, t = 26.5 s.
To solve this problem, we can break it down into three parts:
Step 1: Find the magnitude and direction of the boat's velocity relative to the water.
Step 2: Determine the magnitude and direction of the river's velocity.
Step 3: Combine the boat's velocity relative to the water with the river's velocity to find the boat's velocity relative to the ground.
Let's proceed with the step-by-step solution:
Step 1: Find the magnitude and direction of the boat's velocity relative to the water.
Given:
Boat's speed relative to the water = 9.0 m/s
Angle between the boat's direction and due north = 33º west of north
To find the magnitude of the boat's velocity relative to the water, we can use trigonometry. Since the boat's direction is given relative to the north, we can find the vertical component of the velocity using sin(θ), and the horizontal component using cos(θ).
Vertical component of the velocity (Vv) = 9.0 m/s * sin(33º)
Horizontal component of the velocity (Vh) = 9.0 m/s * cos(33º)
Step 2: Determine the magnitude and direction of the river's velocity.
Given:
River's speed = 4.9 m/s
River flows due east
Since the river flows due east, the velocity vector of the river (Vr) will only have a horizontal component, and no vertical component.
Horizontal component of the river's velocity (Vrh) = 4.9 m/s
Step 3: Combine the boat's velocity relative to the water with the river's velocity to find the boat's velocity relative to the ground.
To find the boat's velocity relative to the ground, we need to add the horizontal components and vertical components of the boat and river velocities.
Vg = Vh + Vrh (horizontal component)
Vg = Vv (vertical component)
Now, we can use the Pythagorean theorem to find the magnitude of the boat's velocity relative to the ground:
Magnitude of the boat's velocity relative to the ground (Vg) = sqrt((Vg)^2 + (Vg)^2)
To find the direction of the boat's velocity relative to the ground, we can use trigonometry. The direction will be the angle from due north, positive if to the east, and negative if to the west.
Direction (θ) = atan(Vv / Vh)
To find the time taken for the boat to cross the river, we can use the formula:
Time = Distance / Velocity
The distance is given as the width of the river (200 m), and the velocity is the horizontal component of the boat's velocity relative to the ground.
Now, let's calculate each part step-by-step:
Step 1:
Vertical component of the boat's velocity (Vv) = 9.0 m/s * sin(33º) ≈ 4.8407 m/s
Horizontal component of the boat's velocity (Vh) = 9.0 m/s * cos(33º) ≈ 7.5220 m/s
Step 2:
Horizontal component of the river's velocity (Vrh) = 4.9 m/s
Step 3:
Magnitude of the boat's velocity relative to the ground (Vg) = sqrt((Vh + Vrh)^2 + Vv^2) ≈ 9.5671 m/s
Direction (θ) = atan(Vv / (Vh + Vrh)) ≈ -29.15º
To find the time taken for the boat to cross the river:
Time = Distance / Velocity
Time = 200 m / (Vh + Vrh) ≈ 17.462 s
Therefore, the solution to the problem is as follows:
(a) The magnitude of the boat's velocity relative to the ground is approximately 9.5671 m/s.
(b) The direction of the boat's velocity relative to the ground is approximately -29.15º (negative if to the west).
(c) It takes approximately 17.462 seconds for the boat to cross the river.
To solve this problem, we need to break it down into smaller components and then combine them to find the final answers. Let's go step by step:
Step 1: Find the velocity of the boat relative to the ground.
The boat's velocity relative to the ground can be found by adding the velocity of the boat relative to the water and the velocity of the water.
Given:
Velocity of boat relative to water (vbw) = 9.0 m/s
Velocity of the river (vr) = 4.9 m/s (due east)
To find the velocity of the boat relative to the ground, we need to use vector addition. We can find the components of the boat's velocity relative to the ground in the north and east directions.
The north component of the boat's velocity relative to the ground (vbnorth) can be found using trigonometry:
vbnorth = vbw * sin(33°)
vbnorth = 9.0 m/s * sin(33°)
The east component of the boat's velocity relative to the ground (vbeast) is simply the velocity of the river:
vbeast = vr
Step 2: Calculate the magnitude and direction of the boat's velocity relative to the ground.
The magnitude of the boat's velocity relative to the ground (vb) can be found using the Pythagorean theorem:
vb = sqrt(vbnorth² + vbeast²)
The direction of the boat's velocity relative to the ground (θ) can be found using trigonometry:
θ = arctan(vbeast / vbnorth)
Step 3: Calculate the time taken to cross the river.
To calculate the time taken to cross the river, we need to divide the width of the river (200 m) by the east component of the boat's velocity relative to the ground (vbeast):
time = 200 m / vbeast
Now let's calculate the results.
Calculations:
vbnorth = 9.0 m/s * sin(33°)
vbeast = 4.9 m/s
vb = sqrt(vbnorth² + vbeast²)
θ = arctan(vbeast / vbnorth)
time = 200 m / vbeast
Solving these equations will give us the answers to the given problem.
Please note that I will now perform these calculations and provide you with the final answers.