A 200-m-wide river flows due east at a uniform speed of 2.0 m/s. A boat with a speed of 8.0 m/s relative to the water leaves the south bank pointed in a direction west of north. What are the (a) magnitude and (b) direction of the boat’s velocity relative to the ground? (c) How long does the boat take to cross the river?
a. Vbg = 8 + 2 = 10m/s.
b. The boat is bointing how many degrees West of North?
To find the boat's velocity relative to the ground, we need to consider the vector addition of the velocity of the boat relative to the water and the velocity of the water.
Let's break down the problem step by step:
(a) To find the magnitude of the boat's velocity relative to the ground, we need to consider the Pythagorean theorem to calculate the resultant velocity.
The boat's velocity relative to the water (Vbw) is 8.0 m/s, and the water's velocity (Vw) is 2.0 m/s due east.
Using the Pythagorean theorem, we can find the magnitude of the resultant velocity (Vrg):
Vrg^2 = Vbw^2 + Vw^2
Substituting the values:
Vrg^2 = (8.0 m/s)^2 + (2.0 m/s)^2
Vrg^2 = 64.0 m^2/s^2 + 4.0 m^2/s^2
Vrg^2 = 68.0 m^2/s^2
Taking the square root of both sides:
Vrg = √(68.0 m^2/s^2)
Vrg ≈ 8.25 m/s
Therefore, the magnitude of the boat's velocity relative to the ground is approximately 8.25 m/s.
(b) To find the direction of the boat's velocity relative to the ground, we can use trigonometry. The direction is given as "west of north."
Since the given direction is relative to the north, we need to find the angle between the boat's velocity vector and the north direction.
Let's assume the angle between the boat's velocity vector and the north direction is θ.
Using trigonometry, we can find the value of θ:
tan(θ) = (Vw)/(Vbw)
tan(θ) = (2.0 m/s) / (8.0 m/s)
tan(θ) = 0.25
Taking the inverse tangent (arctan) of both sides:
θ = arctan(0.25)
θ ≈ 14.04 degrees
Therefore, the direction of the boat's velocity relative to the ground is approximately 14.04 degrees west of north.
(c) To find the time taken for the boat to cross the river, we can use the concept of relative velocity.
The width of the river is given as 200 m, and the boat's velocity relative to the water is 8.0 m/s.
To cross the river, the boat needs to overcome the river's width perpendicular to its flow. We can calculate this distance using basic trigonometry:
Distance = Width of River / sin(θ)
Distance = 200 m / sin(14.04 degrees)
Distance ≈ 827.95 m
Now, we can find the time using the formula:
Time = Distance / Velocity
Time = 827.95 m / 8.0 m/s
Time ≈ 103.49 s
Therefore, it will take approximately 103.49 seconds for the boat to cross the river.