A boat moving at 8.9 km/hr relative to the water is crossing a river 4.6 km wide in which the current is flowing at 2.4 km/hr. At what angle upstream should the boat head to reach a point on the other shore directly opposite the starting point?
how wide has nothing to do with this
Upstream speed = 8.9 sin T - 2.4
if we do not want to go upstream, that must be zero
so
8.9 sin T = 2.4
T = 15.6 degrees
I request the answer
12
Well, if we're talking about a boat crossing a river, I hope the boat knows how to do the river dance! Now, let's figure this out while doing the "flow-y" math dance.
First, we need to break down the boat's velocity into its components. We have the velocity of the boat relative to the water (8.9 km/hr) and the velocity of the current (2.4 km/hr). Since the boat is trying to reach a point directly opposite the starting point, it needs to head at an angle upstream.
Now, let's use some trigonometry to get our groove on. We have a right-angled triangle formed by the boat's velocity relative to the water (hypotenuse), the river's width (opposite side), and the boat's velocity upstream (adjacent side).
To find the angle upstream, we can use the tangent function (tan(angle) = opposite/adjacent). In our case, the opposite side is the width of the river (4.6 km) and the adjacent side is the difference between the boat's velocity relative to the water (8.9 km/hr) and the current's velocity (2.4 km/hr). So, we have tan(angle) = 4.6 / (8.9 - 2.4).
Let's plug in those numbers and calculate that funky angle! Angle = tan^(-1)(4.6 / (8.9 - 2.4)). And the answer is... drumroll, please... the boat needs to head at approximately 31.9 degrees upstream to reach the opposite shore.
So, get your boat, get your dance moves ready, and let's salsa our way across that river!
To solve this problem, we can use vector addition. Let's break down the velocities into their horizontal and vertical components:
- The velocity of the boat relative to the water (Vb) is 8.9 km/hr.
- The velocity of the river current (Vc) is 2.4 km/hr.
Now, let's find the horizontal and vertical components of these velocities using trigonometry.
Horizontal Component of Vb (Vbx):
To find Vbx, we need to use the cosine function. The angle between the boat's velocity and the horizontal direction is the same as the angle we are trying to find. Let's call this angle θ.
Vbx = Vb * cos(θ)
Vertical Component of Vb (Vby):
To find Vby, we use the sine function.
Vby = Vb * sin(θ)
The boat's velocity relative to the water can be decomposed into:
Vb = Vbx + Vby
Now, let's consider the velocity of the boat relative to the ground, which we'll call Vg.
Horizontal Component of Vg (Vgx):
Since the boat is directly opposite its starting point when it reaches the other shore, the horizontal component of Vg should be equal and opposite to the velocity of the river current (Vc).
Vgx = -Vc
Vertical Component of Vg (Vgy):
The vertical component of Vg should be equal to the vertical component of Vb since there is no vertical acceleration.
Vgy = Vby
Now, let's find the resulting velocity of the boat relative to the ground (Vg) by adding the horizontal and vertical components.
Vg = Vgx + Vgy
Next, we need to find the magnitude and direction of Vg. The magnitude will be the speed of the boat relative to the ground.
Magnitude of Vg (speed):
speed = sqrt(Vgx^2 + Vgy^2)
To find the direction, we will use the tangent function. The direction with respect to the horizontal direction will be:
Direction of Vg = arctan(Vgy / Vgx)
Now, let's plug in the given values and calculate the result.