Swimmer and Two Dimensional Equations

  • Thread starter Thread starter wolves5
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on a physics problem involving a woman swimming across a river that is 550 ft wide, flowing at 8 ft/s. When swimming directly across, she is swept downstream, and the distance can be calculated using the time it takes to cross the river. For part (b), by swimming at a 37° angle upstream, the swimmer's velocities must be resolved into components to determine how far downstream she is swept. The correct approach involves using the swimmer's speed of 4 ft/s and applying trigonometric functions to find the components of her velocity.

PREREQUISITES
  • Understanding of vector components in physics
  • Knowledge of basic trigonometric functions (sine and cosine)
  • Familiarity with the concept of relative velocity
  • Ability to apply kinematic equations for motion
NEXT STEPS
  • Learn how to resolve vectors into components using trigonometric functions
  • Study the concept of relative velocity in fluid dynamics
  • Practice kinematic equations for different motion scenarios
  • Explore graphical methods for solving physics problems involving vectors
USEFUL FOR

Students studying physics, educators teaching motion in fluids, and anyone interested in solving real-world problems involving swimming and river currents.

wolves5
Messages
52
Reaction score
0
A river 550 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.

a) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?
d1= ?

b) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight across, as in part (a)), how far downstream is she swept before reaching the opposite bank?
d2 = ?

c) For the conditions of part (b), how long does it take for her to reach the opposite bank?

For this question, I just don't know how to start it. I mean there's no angles. I'm confused because I feel there's not much information like time and all that. I guess I just don't know how to approach this.
 
Physics news on Phys.org
Hey wolves! No worries; I'll help you get started.

wolves5 said:
A river 550 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.

a) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?

Imagine you were the woman, and trying to swim across. How long does it take? You'll be moving along with the water, but that doesn't matter; the river's width doesn't change, so you'll cross in the same amount of time as if the water were still.

You're now an observer on the shore, watching the woman. How far does the water carry her in the time it takes her to reach the other side?

b) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight across, as in part (a)), how far downstream is she swept before reaching the opposite bank?

This is getting a bit more complicated, so you might want to draw a vector diagram of the swimmer's velocity. Then use the same strategy as before: find her velocity perpendicular to the bank, and use that to find how long it takes her to cross. Find her velocity parallel to the bank, and use both that and the time you found to determine how far the river carries her.
 
So for part a, I am using d=vit + 0.5at^2. So, 8(137.5) + 0.5(-9.8)(137.5^2). Is this right? Am I using the right equation?
 
No, because there's no acceleration, and gravity doesn't come into play in this question. Just d=vi*t is all you need.
 
Ok so I got that one down. Now, i don't get part b. What did you mean?
 
If she's swimming at 4ft/s at a 37 degree angle, what's the component of her velocity in the direction perpendicular to the bank? How about the component parallel? (Hint: use sine and cosine)
 
Ok. So, 4sin(37)=2.407 and 4 cos(37)=3.195. Then, I used these velocities and plugged it into D=vt. I used 137.5 as my time. It's still not the right answer.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 30 ·
2
Replies
30
Views
5K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
11K
Replies
5
Views
8K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K