Two canoeists start paddling at the same time and head toward a small island in a lake, as shown in the figure . Canoeist 1 paddles with a speed of 1.60 at an angle of 45 north of east. Canoeist 2 starts on the opposite shore of the lake, a distance of 1.5 due east of canoeist 1.

In what direction relative to north must canoeist 2 paddle to reach the island?

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To determine the direction in which Canoeist 2 must paddle relative to north to reach the island, we need to find the resultant vector of the velocity of Canoeist 1 and the displacement between the two canoeists.

Let's break down the information provided:

- Canoeist 1's velocity: 1.60 m/s at an angle of 45° north of east.
- Canoeist 2's starting position: 1.5 km due east of Canoeist 1.

To find the resultant vector, we can add the displacement vector from Canoeist 2's starting position to the velocity vector of Canoeist 1.

First, let's convert the magnitude and angle of Canoeist 1's velocity into horizontal and vertical components.

Horizontal component (along the east direction):
Vx = V₁ * cos(θ) = 1.60 m/s * cos(45°) = 1.13 m/s

Vertical component (along the north direction):
Vy = V₁ * sin(θ) = 1.60 m/s * sin(45°) = 1.13 m/s

Next, we add the horizontal displacement between the two canoeists to the horizontal component of Canoeist 1's velocity:

Resultant horizontal component:
Rx = Dx + Vx = 1.5 km + 1.13 m/s = 1.5 km + 1.13 m/s

Since the displacement is only in the east direction (x-axis), the vertical component does not change.

Now, we can find the angle that Canoeist 2 needs to paddle relative to north using the tangent function:

tan(θ) = Rx / Vy

θ = arctan(Rx / Vy)

Plugging in the values, we get:

θ = arctan((1.5 km + 1.13 m/s) / 1.13 m/s)

Calculating the arctan, we find:

θ ≈ 53.4°

Therefore, Canoeist 2 must paddle at an angle of approximately 53.4° north of east to reach the island.

To determine the direction relative to north in which canoeist 2 must paddle to reach the island, we need to calculate the vector sum of the velocities of canoeist 1 and canoeist 2.

First, let's represent the given information in vector form:

Velocity of canoeist 1 (v1) = 1.60 m/s at an angle of 45° north of east.
Velocity of canoeist 2 (v2) = unknown magnitue (let's call it v2) at an unknown angle (let's call it θ) relative to north.

Now, we can break down the velocities into their horizontal and vertical components:

v1x = v1 * cos(45°)
v1y = v1 * sin(45°)

v2x = v2 * cos(θ)
v2y = v2 * sin(θ)

Since canoeist 2 starts on the opposite shore of the lake, a distance of 1.5 m due east of canoeist 1, the initial horizontal displacement (Δx) between them is 1.5 m.

By setting up an equation for the horizontal displacement, we can find the time it takes for both canoeists to reach the island:

Δx = v1x * t
1.5 = (1.60 * cos(45°)) * t

Solving for t:
t = 1.5 / (1.60 * cos(45°))

Now, we can calculate the vertical displacement (Δy) using the time (t) we just found:

Δy = v1y * t
Δy = (1.60 * sin(45°)) * (1.5 / (1.60 * cos(45°)))

Finally, to calculate the direction in which canoeist 2 must paddle relative to north, we can use the tangent function:

tan(θ) = Δy / Δx

Substituting the values we found:

tan(θ) = [(1.60 * sin(45°)) * (1.5 / (1.60 * cos(45°))))] / 1.5

Simplifying this equation will give us the tangent of θ, which we can then take the inverse tangent of to find the angle:

θ = arctan([[(1.60 * sin(45°)) * (1.5 / (1.60 * cos(45°))))] / 1.5])

Evaluating this equation will give us the direction relative to north in which canoeist 2 must paddle to reach the island.

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