A boat can travel 2.85 m/s in still water. If the boat points its bow directly across stream whose current is .95 m/s, what is the velocity of the boat relative to the shore?
x speed = .95
y speed = 2.85
speed magnitude = sqrt(.95^2 + 2.85^2)
= 3.00
tan angle to shore = 2.85/.95
A = 71.6 degrees from shoreline
To determine the velocity of the boat relative to the shore, we need to consider the effects of both the boat's speed in still water and the current of the stream. We can use the concept of vector addition to find the resultant velocity.
Let's denote the boat's speed in still water as v_still (2.85 m/s) and the current of the stream as v_current (0.95 m/s).
Since the boat is pointing its bow directly across the stream, the current acts perpendicular to the direction of motion. Thus, we can treat the velocity of the boat relative to the shore as the hypotenuse of a right triangle, with v_still as the base (horizontal component) and v_current as the perpendicular (vertical component).
Using the Pythagorean theorem, we can calculate the magnitude of the boat's velocity relative to the shore:
v_shore = sqrt((v_still)^2 + (v_current)^2)
= sqrt((2.85 m/s)^2 + (0.95 m/s)^2)
= sqrt(8.1225 m^2/s^2 + 0.9025 m^2/s^2)
= sqrt(9.025 m^2/s^2)
≈ 3.00 m/s
Therefore, the velocity of the boat relative to the shore is approximately 3.00 m/s.