A Speedboat whose velocity in still water is 12m/s, is crossing a river flowing due east,at a velocity of 5m/s. If the boat is moving due North, Calculate its velocity across the river and the direction the boat is moving.
draw a diagram. you have a 5-12-13 triangle, right?
tanθ = 5/12
V = 5+12i = 13m/s[67.4o].
To calculate the velocity of the boat across the river, we can use vector addition. The velocity of the boat across the river is the resultant of the boat's velocity in still water and the velocity of the river.
Given:
Velocity of the boat in still water = 12 m/s (due north)
Velocity of the river = 5 m/s (due east)
To find the resultant velocity, we can use the Pythagorean theorem. The magnitude of the resultant velocity is given by:
Resultant velocity = sqrt(Velocity of the boat in still water^2 + Velocity of the river^2)
Plugging in the values:
Resultant velocity = sqrt(12^2 + 5^2)
= sqrt(144 + 25)
= sqrt(169)
= 13 m/s
So, the velocity of the boat across the river is 13 m/s.
To find the direction in which the boat is moving, we can use trigonometry. The angle between the resultant velocity and the north direction is given by:
Angle = arctan(Velocity of the river / Velocity of the boat in still water)
Plugging in the values:
Angle = arctan(5 / 12)
Using a calculator, we find:
Angle ≈ 0.394 radians
Converting radians to degrees:
Angle ≈ 0.394 * (180/π) degrees
Angle ≈ 22.62 degrees
So, the boat is moving at a velocity of 13 m/s across the river, in a direction approximately 22.62 degrees east of north.
To solve this problem, we can use vector addition. We need to find the resultant velocity of the boat, which is the combination of its velocity in still water and the velocity of the river.
First, let's draw a diagram to visualize the problem. You can draw a coordinate system with the x-axis representing the east-west direction and the y-axis representing the north-south direction. The boat's velocity in still water (12 m/s) is directed north, so it lies along the positive y-axis. The river's velocity (5 m/s) is directed east, so it lies along the positive x-axis.
The velocity across the river is the component of the resultant velocity along the x-axis, and the direction of the boat's movement is given by the angle it makes with the positive x-axis.
To find the velocity across the river, we can use the Pythagorean theorem. Let's call the velocity across the river Vx and the resultant velocity V.
V^2 = (Vx)^2 + (Vy)^2
Since we know the velocity in still water is directed north, Vy is equal to the boat's velocity in still water, which is 12 m/s.
V^2 = (Vx)^2 + (12)^2
To find the value of Vx, we need to note that the velocity of the river is acting perpendicular to the boat's velocity in still water, forming a right triangle. So, we have:
tan(theta) = (opposite/adjacent)
tan(theta) = Vx/5
Vx = 5 * tan(theta)
Now we can substitute this expression into the Pythagorean theorem equation:
V^2 = (5 * tan(theta))^2 + (12)^2
Since we want to find V, we can solve this equation for V by taking the square root of both sides:
V = sqrt((5 * tan(theta))^2 + (12)^2)
To find the direction the boat is moving, we need to find the angle theta. We can use the arctan function:
tan(theta) = Vx/5
theta = arctan(Vx/5)
Now that we have the value of theta, we can calculate it using a scientific calculator or specialized software. Substituting the value of theta into the equation for V will give us the velocity across the river.
Note: Keep in mind that angles can be measured in degrees or radians, so make sure you are consistent with your calculator settings and the units you use.