Calculating Horse's Average Velocity: 130m in 18.0 s

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Homework Help Overview

The problem involves calculating the average velocity of a horse that moves away from its trainer and then returns partway back. The horse travels 130 meters in 18.0 seconds and then gallops halfway back in 5.2 seconds. The context is centered around understanding average velocity as a vector quantity, considering direction and distance.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need for the original poster to show their work and apply the relevant equations. There are attempts to calculate average velocity using distances and times, but confusion arises regarding the signs of the velocities and the correct application of the concept of average velocity.

Discussion Status

Some participants have provided guidance by suggesting that the original poster think through the problem using a simpler analogy involving driving. There is an acknowledgment of the need to clarify the distinction between speed and velocity, as well as the importance of direction in the calculations. The discussion reflects a mix of interpretations and attempts to understand the problem better.

Contextual Notes

The original poster has indicated limited experience with physics, having only taken the subject for a few days, which may contribute to their difficulty in grasping the concepts being discussed.

lostfan176
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Homework Statement



A horse canters away from its trainer in a straight line, moving 130 m away in 18.0 s. It then turns abruptly and gallops halfway back in 5.2 s. Average speed is 8.41 m/s
Calculate its average velocity for the entire trip, using "away from the trainer" as the positive direction.

Homework Equations



velocity equation and speed equation
 
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Welcome to the PF, lostfan. You need to show us some of your own work, before we can offer tutorial advice. Them's the PF rules.

So please show us the equations that you allude to, and take a try at applying them to this problem.
 
well i tried 120 - (and +) 65 over 18 - (and +) 5.2

i kept getting the wrong answers ( i check if its wrong or right cause I am using webassign.net)
 
lostfan176 said:
well i tried 120 - (and +) 65 over 18 - (and +) 5.2

i kept getting the wrong answers ( i check if its wrong or right cause I am using webassign.net)

Sorry, I'm not tracking what you are saying. The positive velocity for the first part is calculated based on 130m in 18s, right? What is that? And the negative velocity for the second part is what? And remember that the total average will be whatever velocity vector would have gotten you to that same final spot in the same total amount of time...
 
i don't understand what you are saying sorry i only took phys for 3 days now
 
lostfan176 said:
i don't understand what you are saying sorry i only took phys for 3 days now

Let's take a simpler example. Let's say you drive a car to the right (call that the + direction) at 30mph for an hour, and then turn around and drive back half the way at that same 30mph (which would take half an hour, right?).

Velocity is a vector with speed (the scalar part) and a direction. So the velocity for the first part of the trip was +30mph for 60 minutes, and the velocity vector for the second part was -30mph for 30 minutes. Now think about what the average velocity vector would be... It's the vector that would get you from your starting place to the finishing place in 90 minutes, right?

Now see if that helps your thinking about the problem that you are working on. Just spend some time thinking it through. And review the parts of your textbook that talk about vectors and velocity and stuff like that.

I have to bail here pretty soon, but if you still have problems after a while, post more of your work and you should get some help from one of the other folks.
 
thx for helping but i don't think its goin at a constant speed
 
lostfan176 said:
thx for helping but i don't think its goin at a constant speed

The way the problem is worded, the speed is one constant value for the outbound portion, and a different constant speed for the inbound portion. I think you have what you need to figure it out. Good luck!
 

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