a student stands on the southern bank of a river that is 50m wide and has a current with a velocity of 1.1m/s [E]. the student needs to get directly across the river using their boat. if the boat's motor can propel the boat forwards with a speed of 3.8m/s, in what direction should the student point their boat to ensure they travel directly across the river to their destination?
To ensure that the student travels directly across the river, they need to aim their boat slightly upstream from their desired destination.
First, let's break down the velocities involved:
1. Velocity of the river current: 1.1 m/s [E] (east)
2. Velocity of the boat's motor: 3.8 m/s
Now we need to determine the angle at which the student should point their boat to ensure they travel directly across the river.
Let's assume the angle as θ, measured clockwise from the east.
Using vector addition, we can find the resultant velocity (v_res) of the boat:
v_res = v_boat + v_river
v_boat = 3.8 m/s at an angle of θ
v_river = 1.1 m/s [E]
To find v_res, we can resolve v_boat into its horizontal component (v_h) and vertical component (v_v):
v_h = v_boat * cos(θ)
v_v = v_boat * sin(θ)
Therefore, the horizontal component of the resultant velocity is:
v_res_h = v_h + v_river = v_boat * cos(θ) + v_river
The vertical component of the resultant velocity is:
v_res_v = v_v = v_boat * sin(θ)
To ensure that the student travels directly across the river, the vertical component (v_v) of the resultant velocity must be zero. This means that the boat should not have any vertical velocity, only horizontal velocity.
Setting v_res_v = 0, we have:
v_boat * sin(θ) = 0
Since sin(θ) = 0 when θ = 0° or θ = 180°, we have two possible solutions:
1. θ = 0°: This means the boat should point directly east, parallel to the river current.
2. θ = 180°: This means the boat should point directly west, opposite to the river current.
However, pointing the boat directly east will cause the river current to push the boat downstream, away from the desired destination. Therefore, the correct angle for the boat to ensure it travels directly across the river is θ = 180°, i.e., the boat should point directly west, opposite to the river current.
To ensure they travel directly across the river to their destination, the student needs to account for the velocity of the river current while pointing their boat. Here are the steps to determine the direction in which the student should point their boat:
Step 1: Draw a diagram: Draw a diagram to visualize the problem. Label the river's width as 50m and the current's velocity as 1.1m/s [E]. Also, indicate the boat's motor speed as 3.8m/s.
Step 2: Break velocities into components: Break down the velocities into their x and y components. The river current velocity is purely in the x-direction, so its x-component is 1.1m/s, and its y-component is 0m/s. The boat's motor speed is purely in the forward direction, so its x-component is 3.8m/s, and its y-component is 0m/s.
Step 3: Determine the resultant velocity: To travel directly across the river, the boat's resultant velocity must have an x-component equal to the width of the river (50m), and a y-component equal to zero. Let's assume the angle at which the boat is pointed relative to the x-axis is θ.
Since the resultant velocity's x-component is equal to the sum of the river's current and the boat's x-component, we have:
3.8cosθ + 1.1 = 50
Step 4: Solve for θ: Rearrange the equation to solve for θ:
3.8cosθ = 50 - 1.1
cosθ = (50 - 1.1) / 3.8
Step 5: Calculate θ: Use inverse cosine (cos^(-1)) to calculate θ:
θ = cos^(-1) [(50 - 1.1) / 3.8]
Step 6: Calculate the result: Use a calculator to find the value of θ, which gives the direction the student should point their boat to travel directly across the river.
Note: The calculation provides an angle measured clockwise from the positive x-axis.
To determine the direction in which the student should point their boat, we need to consider the velocity of the boat and the velocity of the current.
Let's break down the velocities involved:
1. Velocity of the current: The current is flowing with a speed of 1.1 m/s to the east (towards the direction of the positive x-axis), which means it has a velocity vector of 1.1 m/s [E].
2. Velocity of the boat: The boat's motor can propel it forwards with a speed of 3.8 m/s (magnitude) relative to still water. This means that when there is no current, the boat can move in any direction at a speed of 3.8 m/s.
Now, we want to find the direction in which the boat should be pointed so that it cancels out the effect of the current and allows the boat to move directly across the river.
To achieve this, we can use vector addition. We want the resultant velocity vector to be directly across the river, which means it should have no component in the east-west direction (no x-component). Therefore, we need to cancel out the x-component of the current's velocity with the boat's velocity.
Let's break down the velocity vectors into their x (east-west) and y (north-south) components:
1. Velocity of the current: v_current = 1.1 m/s [E]
- x-component: v_current_x = 1.1 m/s
- y-component: v_current_y = 0 m/s
2. Velocity of the boat: v_boat = 3.8 m/s [?]
- x-component: v_boat_x = 3.8 m/s * cos(θ) (where θ is the angle between the boat's direction and the positive x-axis)
- y-component: v_boat_y = 3.8 m/s * sin(θ) (where θ is the angle between the boat's direction and the positive x-axis)
Now, since we want the resultant velocity to have no x-component (directly across the river), we need to make v_current_x + v_boat_x = 0. Therefore:
1.1 m/s + 3.8 m/s * cos(θ) = 0
Solving for cos(θ):
cos(θ) = -1.1 m/s / 3.8 m/s
θ = arccos(-1.1 m/s / 3.8 m/s)
Using a calculator, we find that θ ≈ 107.06°.
Therefore, the student should point their boat at an angle of approximately 107.06° (measured counterclockwise from the positive x-axis, or east) to ensure they travel directly across the river to their destination.