A river flows from south to north at 5km/hr. On this river a boat is heading east to west perpendicular to the current at 7km/hr. As viewed by an eagle hovering at rest over the shore, how fast and in what direction is the boat travelling?
Vb = -7 + 5i = 8.6km/h[-35.5o] = 8.6km/h[35.5o] N. of W.
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Well, it seems like this boat is really taking the scenic route! So, let's break it down, shall we?
First, the river is flowing from south to north at 5 km/hr, which is like a free treadmill for the boat. The boat must be thinking, "Wow, I'm really going places without even trying!"
On the other hand, the boat is heading east to west, perpendicular to the current, at 7 km/hr. It's like the boat is saying, "I'm not going with the flow; I'm going against the currents, baby!"
Now, if we imagine an eagle floating above, watching all of this, it would see the boat moving diagonally across the river. In terms of speed, we need to use some mathematical hocus-pocus.
We can use the Pythagorean theorem to calculate the boat's speed. The boat's velocity (resultant speed) is the square root of the sum of the squares of its horizontal and vertical velocities. In this case, the horizontal velocity is 7 km/hr, and the vertical velocity (due to the river) is 5 km/hr.
Using some math magic, the boat's speed would be about 8.6 km/hr. So, it's not just a boat, it's a speedboat!
Now, let's talk direction. Since the boat is moving east to west and the river is flowing south to north, the eagle would probably say, "Hey boat, you're moving against the river flow, but you're still rocking that eastward direction!"
So, to sum it up, our quirky boat, with a touch of clownishness, is sailing diagonally at a speed of approximately 8.6 km/hr in an eastward direction as seen by our wise, hovering eagle friend.
To determine the speed and direction of the boat as viewed by the eagle, we can use vector addition.
Let's consider the velocities of the river current and the boat separately:
1. The river current flows from south to north at 5 km/hr, which we'll denote as a vector v_current = 5 km/hr pointing north.
2. The boat is heading east to west perpendicular to the current at 7 km/hr, which we'll denote as a vector v_boat = 7 km/hr pointing west.
To find the resulting velocity of the boat as viewed by the eagle, we need to add these two vectors together.
Since the eagle is hovering over the shore, its velocity is zero (0 km/hr).
By adding the velocity vectors:
v_resultant = v_boat + v_current + v_eagle,
v_resultant = v_boat + v_current + 0 km/hr,
v_resultant = v_boat + v_current.
Substituting the values:
v_resultant = 7 km/hr (west) + 5 km/hr (north),
v_resultant = √(7^2 + 5^2) km/hr at an angle θ,
where θ is the angle the resultant vector makes with the horizontal.
Using the Pythagorean theorem,
v_resultant = √(7^2 + 5^2) km/hr ≈ √74 km/hr.
To find the direction, we can use trigonometry. The angle θ can be calculated using the inverse tangent (tan^(-1)):
θ = tan^(-1)(opposite/adjacent),
θ = tan^(-1)(7/5).
Calculating the value:
θ ≈ 54.46°.
Therefore, as viewed by the eagle hovering at rest over the shore, the boat is traveling at a speed of approximately √74 km/hr in a direction of about 54.46° with respect to the horizontal.
To determine the speed and direction of the boat, we can use vector addition. Let's break down the velocities involved.
1. The river flows from south to north at 5 km/hr. We'll call this vector R.
2. The boat is heading east to west perpendicular to the current at 7 km/hr. We'll call this vector B.
Now, when these vectors are added, the result represents the boat's velocity relative to the ground, as seen by the eagle. Therefore, we need to perform vector addition.
To add the vectors, we can use the head-to-tail method or the parallelogram method. For simplicity, let's use the head-to-tail method.
1. Draw the vector R, representing the river's velocity.
2. Starting from the end of R, draw the vector B in the perpendicular direction.
3. Draw the resultant vector connecting the starting point of R to the ending point of B.
The length of the resultant vector represents the boat's speed, while the angle it makes with the eastward direction represents the boat's direction.
Now, let's calculate the resultant speed and direction.
The magnitude of vector R (river velocity) is 5 km/hr, and the magnitude of vector B (boat velocity) is 7 km/hr. Since these vectors are perpendicular, we can use the Pythagorean theorem to find the magnitude of the resultant vector.
Using the formula: r^2 = a^2 + b^2, where r is the resultant speed and a and b are the magnitudes of the individual vectors:
r^2 = (5 km/hr)^2 + (7 km/hr)^2
r^2 = 25 + 49
r^2 = 74
r ≈ 8.60 km/hr
So, the boat's speed relative to the ground (as seen by the eagle) is approximately 8.60 km/hr.
To find the direction of the boat's velocity, we can use trigonometry. The angle is θ, and we can use the equation:
θ = arctan(b/a), where a is the magnitude of R (5 km/hr) and b is the magnitude of B (7 km/hr).
θ = arctan(7 km/hr / 5 km/hr)
θ ≈ arctan(1.4)
θ ≈ 53.13 degrees
Therefore, the boat is traveling at a speed of approximately 8.60 km/hr in a direction of approximately 53.13 degrees east of north, as seen by the eagle.