Average speed & average velocity Problem

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Homework Help Overview

The discussion revolves around a problem involving average speed and average velocity in the context of a mountain climbing scenario. The original poster presents a series of calculations related to the ascent and descent of a climber, including the determination of average angle of ascent, average speed, and average vertical velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the appropriate equations for calculating the average angle of ascent and average speeds, with some confusion about the definitions of vertical velocity and the correct use of trigonometric functions.

Discussion Status

Some participants have provided hints and clarifications regarding the geometric interpretation of the problem, while others are exploring the relationships between different variables. There is an ongoing exchange of ideas about the calculations and the correct application of formulas without reaching a final consensus.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available and the methods they can use. There is also a focus on ensuring that answers are expressed in the correct units, such as meters per second.

davidco
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Homework Statement


Brandon is climbing a mountain. He starts from an elevation of H1=511m above sea level. In 6.18 hours he climbs to an elevation of H2=2746m above sea level. The length of his path is L1=16000m.

A)What is the average angle of ascent the path makes with the horizontal?Theta=
B)What is his average speed along the path in meters/second? average speed=
C)What is his average vertical velocity? Vup=
D)After he eats he hikes back down to their starting point in T2=4.24 hours
What is his average vertical velocity? Vdown=

Homework Equations


avg speed=total distance/total time
avg velocity=v=Delta x/delta t
tan theta=opposite/adjacent


The Attempt at a Solution



A)tan theta=opposite/adjacent
Im so confused as to what equations to use.

B)avg speed= total distance/total time

16000M/22680S=0.705m/s


C)avg velocity=v=Delta x/delta t

D)Confused on D as well

Please give some suggestions or the correct equations will be great.thanx Dave
 
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You are on the right track. If you haven't done it already, one thing that helps is to draw a picture.

For a, you have the correct equation. Hint: what does L1 represent in a simple idealized geometric shape?

For c & d: be careful. The problem says average vertical velocity.

Also, looks like you double posted by mistake.:smile:
 
Last edited:
Hotvette thanks a lot for the help just got a few more questions.

If your talking in terms of a triangle L1 would represent my hypotenuse I believe. So would 16000m be my hypotenuse is what your suggesting. But don't I have to choose the equation involving the hypotenuse to find the angle. Like:

sin theta= opp/hyp
or
cos theta= adj/hyp

But my question is what is the equation for vertical velocity?

Thanx Vette
 
For the angle yes, use:

sin theta = opp/hyp

average vertical velocity = vertical displacement/time
 
Thank you learn,


But I am wondering if my hyp=16000m and I am trying to find theta. what's my opp or how do i find my opp.?
 
davidco said:
Thank you learn,


But I am wondering if my hyp=16000m and I am trying to find theta. what's my opp or how do i find my opp.?

What distance does he travel vertically? they give you initial height and final height...
 
Thanx Learn,



Ok so if I subtract the H2=2746m- 511m=my distance is 2235m

So to find theta I did this:

sin theta=opp/hyp

sin theta=2235m/16000
theta=7.98 degrees or 8 degrees

Please tell me this is correct
 
davidco said:
Thanx Learn,



Ok so if I subtract the H2=2746m- 511m=my distance is 2235m

So to find theta I did this:

sin theta=opp/hyp

sin theta=2235m/16000
theta=7.98 degrees or 8 degrees

Please tell me this is correct

yup, looks right to me. But I get 8.03 degrees.
 
Thanx,

For my average speed =total distance/total time,do I use the 16000m for total distance/6.18hrs or do i convert hrs to seconds. Since my answer has to be in m/s?
 
  • #10
Yeah, I think converting to s is best. otherwise your answer will be in m/hr... best to use m/s.
 
  • #11
Question so if vertical speed= deltaV/T,would my deltaV= .705m/s?I got .705m/s from my average velocity equation which was 16000m/22680s=.705m/s for avgV. Is average vertical velocity and vertical velocity the same equation?
 
  • #12
vertical velocity is vertical displacement which you know is 2235m divided by time.
 

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