A girl swims across a river. When she swims in still water, she swims at 1.25 m/s, the river flows parallel to its banks at v m/s. The girl aims to swim upstream at angle θ degree to the river bank so that her resultant velocity of 1 m/s is along AB, perpendicular to the river bank.

Find V and θ

velocity downstream is zero (she swims directly across)

1.25cosTheta=1 solve for theta

1.25sinTheta=V solve for V

is the same thing if in problem say

"at angle θ degrees parallel to the river bank"

To find the value of V and θ, we can use vector addition and trigonometry. Let's break down the problem step by step:

Step 1: Draw a diagram
Draw a diagram representing the river, girl's swim path, and the velocity components.

A
-----------------------
| θ |
| |
-->|-------->VB---------


|
V

The river flows from left to right at velocity VB. The girl aims to swim at an angle θ degree to the river bank, which we will call VA.

Step 2: Break down the velocity components
To solve the problem, we need to break down the girl's velocity into its components: one component parallel to the river flow (VB) and another component perpendicular to the river flow (VA).

The velocity VA is perpendicular to the river and is equal to the desired resultant velocity (1 m/s) since it is along AB. Therefore, VA = 1 m/s.

The velocity VB is parallel to the river flow. It can be calculated using the Pythagorean theorem:

VB = √(VBx² + VBy²)

We know that VBx (the x-component of VB) is equal to the girl's swimming speed in still water (1.25 m/s) because it is the component parallel to the river flow. So, VBx = 1.25 m/s.

Now, we need to find VBy (the y-component of VB) by using trigonometry. We can write:

VBy = VB * sin(θ)

Step 3: Solve for V and θ
To find V and θ, we need to solve for VB and θ in terms of the given information.

First, substitute the given values into the equation for VB:

VB = √(VBx² + VBy²)
= √((1.25 m/s)² + (VB * sin(θ))²)

Now, rearrange the equation to isolate VB:

VB² = (1.25 m/s)² + (VB * sin(θ))²

VB² - (VB * sin(θ))² = (1.25 m/s)²

VB² (1 - sin²(θ)) = (1.25 m/s)²

VB² = (1.25 m/s)² / (1 - sin²(θ))

VB = √((1.25 m/s)² / (1 - sin²(θ)))

Now, to solve for θ, we use the equation for VBy:

VBy = VB * sin(θ)
VB * sin(θ) = VB * sin(θ)

Therefore, θ can take any value since it cancels out on both sides.

To summarize:
- VB = √((1.25 m/s)² / (1 - sin²(θ)))
- θ can take any value.

Note: When plugging in values for θ, make sure to convert the angle to radians if necessary.