A boat crosses a river of width 123 m in which
the current has a uniform speed of 1.49 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 2.66 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.
heading (i.e., the direction in which the boat points)
You steer on a heading.
You take a bearing on something else.
Any way:
speed component across = 1.49 m/s
speed component downstream = 2.66
the components are perpendicular to each other so
magnitude of speed = sqrt(1.49^2+2.66^2)
= sqrt( 2.22+7.08)
= 3.05 m/s
People who write math texts apparently do not go to sea.
To find the magnitude of the speed of the boat relative to a stationary shore observer, we can use vector addition.
Let's break down the vectors involved:
1. The velocity of the boat relative to the water: This is given as 2.66 m/s. Let's call it vector B.
2. The velocity of the current: This is given as 1.49 m/s. Let's call it vector C.
3. The velocity of the boat relative to the stationary shore observer: Let's call it vector A. This is what we want to find.
Since the boat is moving perpendicular to the river, we can treat the boat's velocity and the current's velocity as perpendicular vectors. Therefore, we can use the Pythagorean theorem to find the magnitude of their combined velocity.
Using Pythagorean theorem:
(A^2) = (B^2) + (C^2)
Plugging in the given values:
(A^2) = (2.66^2) + (1.49^2)
(A^2) = 7.0756 + 2.2201
(A^2) = 9.2957
Taking the square root of both sides:
A = sqrt(9.2957)
A = 3.052 m/s
Therefore, the magnitude of the speed of the boat relative to a stationary shore observer is 3.052 m/s.