A 410-m-wide river has a uniform flow speed of 2.0 m/s through a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of 7.9 m/s with respect to the water. There is a clearing on the north bank 36 m upstream from a point directly opposite the clearing on the south bank. (a) At what angle, measured relative to the direction of flow of the river, must the boat be pointed in order to travel in a straight line and land in the clearing on the north bank? (b) How long will the boat take to cross the river and land in the clearing?

To answer the given questions, we need to use the concept of vector addition and relative velocities.

(a) To determine the angle at which the boat must be pointed, we can use the concept of relative velocities. The boat's velocity relative to the water is given as 7.9 m/s, and the river's flow speed is 2.0 m/s towards the east.

The boat needs to travel in a straight line across the river to land in the clearing on the north bank. This means the boat should have a resultant velocity that combines its velocity relative to the water and the velocity of the river.

To obtain the required angle, we can consider the triangle formed by the boat's velocity relative to the water, the river's velocity, and the resultant velocity.

Let's consider the horizontal component in the direction of the river's flow and the vertical component perpendicular to the river's flow.

Horizontal component of the boat's velocity relative to the water = 7.9 m/s
River's velocity = 2.0 m/s (opposite direction, since it flows towards the east)

Now, if the boat is pointed directly across the river, the horizontal component of the resultant velocity would be the sum of the horizontal components of the boat's velocity relative to the water and the river's velocity.

So, horizontal component of the resultant velocity = 7.9 m/s - 2.0 m/s = 5.9 m/s

The vertical component of the resultant velocity should be zero to ensure the boat travels in a straight line across the river.

Using the Pythagorean theorem, we can find the magnitude of the resultant velocity:

Resultant velocity = √((horizontal component of resultant velocity)^2 + (vertical component of resultant velocity)^2)

Since the vertical component should be zero, we have:

Resultant velocity = √((5.9 m/s)^2 + 0^2) = 5.9 m/s

Now, to determine the angle, we can use trigonometry.

tan(angle) = (vertical component of resultant velocity) / (horizontal component of resultant velocity)

Since the vertical component is zero:

tan(angle) = 0 / (5.9 m/s) = 0

Therefore, the angle at which the boat must be pointed is 0 degrees (or in the direction of the river's flow).

(b) To find the time it takes for the boat to cross the river and land in the clearing, we need to calculate the distance it needs to cover.

The north bank clearing is located 36 meters upstream from a point directly opposite the clearing on the south bank.

The total distance to be covered is the width of the river plus the additional distance upstream.

Total distance = width of the river + 36 m = 410 m + 36 m = 446 m

Since we know the boat's velocity relative to the water is 7.9 m/s, we can use the formula: time = distance / velocity

Time = 446 m / 7.9 m/s

Calculating this, we find:

Time ≈ 56.5 seconds

Therefore, it will take approximately 56.5 seconds for the boat to cross the river and land in the clearing.

To solve this problem, we can break it down into two components: the horizontal motion (across the river) and the vertical motion (downstream).

(a) To find the angle, we can use the concept of vector addition. The boat's velocity relative to the water is its actual speed (7.9 m/s) in the forward direction. The river's velocity is its speed (2.0 m/s) in the east direction. The resultant velocity (velocity of the boat relative to the ground) should be the vector sum of these two velocities, pointing directly towards the north bank clearing.

Let's assume the angle of the boat's heading with respect to the river flow is θ.

Horizontal Component:
The boat's horizontal component of velocity is given by V_horizontal = V_boat * cos(θ).

Vertical Component:
The river's velocity is acting against the boat, giving rise to a vertical component of velocity. The boat's vertical component of velocity is given by V_vertical = V_river + V_boat * sin(θ).

Since the boat is moving in a straight line towards the clearing on the north bank, the vertical component of velocity should be zero. Hence, we have:

V_river + V_boat * sin(θ) = 0

We can substitute the given values:
2.0 m/s + 7.9 m/s * sin(θ) = 0

Solving this equation will give us the value of θ.

(b) To find the time taken to cross the river, we can use the horizontal component of velocity, as the boat moves directly across the river.

Time = Distance / Velocity
Time = 410 m / (7.9 m/s * cos(θ))

We can substitute the previously calculated value of θ to find the time taken.

Let's calculate the answers to these questions now.

a. d = -36 + 410, Q2.

Tan A = 410/-36 = -11.38889
A = -85o = N. of W. = 95o CCW = 5o W of N.

b. d = sqrt(36^2+410^2) = 412 m.

d = V*t
t = d/V = 412/7.9 = 52 s.