The speed of a powerboat in still water is 35 mi/h. It is traveling on a river that flows directly south at 8 mi/h.the boat heads directly west across the river. What are the resulting speed and direction of the boat?
I tried this problem and got approximately 35.8 mph and 12.9 degrees south of west. I hope this is right.
To find the resulting speed and direction of the boat, we can use vector addition.
Let's consider the velocity of the boat in still water as Vboat and the velocity of the river flow as Vriver.
Given:
Speed of the boat in still water (Vboat) = 35 mi/h
Speed of the river flow (Vriver) = 8 mi/h
To find the resulting speed and direction, we need to add these two vectors:
Vresultant = Vboat + Vriver
Now let's break down the vectors into their components:
Vboat = 35 mi/h (westward)
Vriver = 8 mi/h (southward)
The westward component of the boat's velocity is 35 mi/h, and the southward component of the river's velocity is 8 mi/h. Since these components are perpendicular, we can use the Pythagorean theorem to find the magnitude of the resultant velocity:
R = sqrt((Vx)^2 + (Vy)^2)
Where Vx is the east-west component and Vy is the north-south component of the resultant velocity.
Plugging in the values:
R = sqrt((35)^2 + (8)^2)
R = sqrt(1225 + 64)
R = sqrt(1289)
R ≈ 35.93 mi/h (resulting speed of the boat)
To find the direction of the boat, we need to find the angle it makes with the westward direction. We can use trigonometry to calculate this angle:
θ = arctan(Vy/Vx)
Where Vy is the north-south component and Vx is the east-west component of the resultant velocity.
Plugging in the values:
θ = arctan(8/35)
θ ≈ 12.06° (resulting direction of the boat)
Therefore, the resulting speed of the boat is approximately 35.93 mi/h in a direction approximately 12.06° westward across the river.
To determine the resulting speed and direction of the boat, we can use vector addition. We'll break down the boat's motion into its horizontal (west-east) and vertical (north-south) components.
Given:
Speed of the powerboat in still water (in the absence of river current) = 35 mi/h
Speed of the river current flowing south = 8 mi/h
Horizontal Component:
Since the boat is heading directly west across the river, its horizontal speed is in the west direction. So, the horizontal component of the boat's velocity is 35 mi/h towards the west.
Vertical Component:
Due to the river current flowing directly south, the boat experiences a vertical velocity toward the south with a magnitude of 8 mi/h.
Now, we can find the resulting speed and direction of the boat by vectorially adding the horizontal and vertical components of its velocity.
Speed:
Using the Pythagorean theorem, we can find the resulting speed:
Resultant Speed = √(Horizontal Component^2 + Vertical Component^2)
= √((35 mi/h)^2 + (8 mi/h)^2)
≈ √(1225 mi^2/h^2 + 64 mi^2/h^2)
≈ √1289 mi^2/h^2
≈ 35.93 mi/h (rounded to two decimal places)
Direction:
The direction of the boat can be found using trigonometry. We have a right-angled triangle, where the horizontal component is the base, the vertical component is the height, and the resulting velocity is the hypotenuse.
Tan(θ) = Vertical Component / Horizontal Component
θ = atan(Vertical Component / Horizontal Component)
= atan(8 mi/h / 35 mi/h)
≈ atan(0.23)
≈ 13.2° (rounded to one decimal place)
Therefore, the resulting speed of the boat is approximately 35.93 mi/h, and its direction is approximately 13.2° west of south.