A boat crosses a river of width 98.4 m
in which the current has a uniform speed of 1.84 m/s. The pilot maintains a bearing (i.e.,the direction in which the boat points) perpendicular to the river and a throttle settingto give a constant speed of 2.11 m/s relative
to the water.
What is the magnitude of the speed of the boat relative to a stationary shore observer?
Answer in units of m/s.
High school texts --- BOO !
HEADING is the direction you point a boat or plane
BEARING is the angle you measure to a lighthouse FROM your boat or plane.
That said:
V across = 2.11
V down stream = 1.84
|V| = sqrt (2.11^2 + 1.84^2)
|V| = 2.8 m/s
To find the magnitude of the speed of the boat relative to a stationary shore observer, we need to consider the effect of both the boat's speed relative to the water and the current of the river.
Let's break down the problem into two components: the horizontal component and the vertical component.
1. Horizontal Component:
Since the pilot maintains a bearing perpendicular to the river, the horizontal component of the boat's speed relative to the stationary shore observer will be equal to the speed of the boat relative to the water. Therefore, the horizontal component is 2.11 m/s.
2. Vertical Component:
The current of the river has a uniform speed of 1.84 m/s. As the boat crosses the river, the current will push the boat downstream. The vertical component of the boat's speed relative to the stationary shore observer is equal to the speed of the current.
Now, we can use these two components to find the magnitude of the speed of the boat relative to the stationary shore observer using the Pythagorean theorem:
Speed^2 = Horizontal Component^2 + Vertical Component^2
Speed^2 = (2.11 m/s)^2 + (1.84 m/s)^2
Speed^2 = 4.4521 m^2/s^2 + 3.3856 m^2/s^2
Speed^2 = 7.8377 m^2/s^2
Taking the square root of both sides, we get:
Speed = √7.8377 m^2/s^2
Speed ≈ 2.7999 m/s
Therefore, the magnitude of the speed of the boat relative to the stationary shore observer is approximately 2.7999 m/s.