A boat crosses a river of width 164 m in which
the current has a uniform speed of 1.3 m/s.
The pilot maintains a bearing (i.e., the direction
in which the boat points) perpendicular
to the river and a throttle setting to give a
constant speed of 2.4 m/s relative to the water.
What is the magnitude of the speed of the
boat relative to a stationary shore observer?
v=_/(1.3^2 + 2.4^2)
v = 2.72
V = sqrt(x^2+y^2).
X = 1.3 m/s, Y = 2.4 m/s, V = ?.
To find the magnitude of the speed of the boat relative to a stationary shore observer, we can use vector addition.
Let's denote the speed of the boat relative to the water as VB (velocity of the boat) and the speed of the current as VC (velocity of the current).
The magnitude of the speed of the boat relative to a stationary shore observer is given by the magnitude of the resultant velocity vector.
Using vector addition, we have:
VB^2 = VC^2 + V^2
Where V represents the magnitude of the velocity of the boat relative to the stationary shore observer.
Given:
VB = 2.4 m/s (speed of the boat relative to the water)
VC = 1.3 m/s (speed of the current)
Plugging in the values:
V^2 = 1.3^2 + 2.4^2
V^2 = 1.69 + 5.76
V^2 = 7.45
V = square root of 7.45
V ≈ 2.73 m/s
Therefore, the magnitude of the speed of the boat relative to a stationary shore observer is approximately 2.73 m/s.
To find the magnitude of the speed of the boat relative to a stationary shore observer, we need to consider the vector components of the boat's motion.
Let's break down the motion of the boat into horizontal and vertical components.
1. Horizontal Component:
Since the boat maintains a bearing perpendicular to the river, its motion in the horizontal direction is solely due to the river's current. The magnitude of the horizontal component of the boat's velocity is given by the speed of the river's current, which is 1.3 m/s.
2. Vertical Component:
The boat's motion in the vertical direction is determined by its speed relative to the water. The magnitude of the vertical component of the boat's velocity is the same as its speed relative to the water, which is 2.4 m/s.
Now, let's find the magnitude of the boat's velocity relative to the stationary shore observer using the Pythagorean theorem:
v = √(vh^2 + vv^2)
where v is the magnitude of the boat's velocity, vh is the horizontal component of the boat's velocity, and vv is the vertical component of the boat's velocity.
Plugging in the values we know:
vh = 1.3 m/s
vv = 2.4 m/s
v = √(1.3^2 + 2.4^2)
v = √(1.69 + 5.76)
v = √7.45
v ≈ 2.73 m/s
Therefore, the magnitude of the speed of the boat relative to a stationary shore observer is approximately 2.73 m/s.