Calculating Average Velocity from Position vs Time Graph

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Homework Help Overview

The discussion revolves around calculating average velocity from a position versus time graph for a particle moving along the x-axis. Participants are tasked with finding average velocities over specified time intervals based on the provided graph.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for average velocity, questioning the application of the equation x(t_2) - x(t_1) / t_2 - t_1. There is confusion regarding the interpretation of x(t_2) and the values to be used in the calculation.

Discussion Status

Some participants are attempting to clarify the formula and the values they should use, while others express uncertainty about their calculations. There is an ongoing exploration of the correct interpretation of the average velocity formula, and some guidance has been offered regarding the proper substitution of values.

Contextual Notes

Participants are working with a graph that is not visible in the thread, which may lead to misunderstandings about the position values at specific times. There is also a mention of potential rounding issues affecting the answers provided.

chanv1
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Homework Statement



The position versus time for a certain particle moving along the x-axis is shown in the figure below.

http://img87.imageshack.us/img87/6101/physicswg4.th.jpg http://g.imageshack.us/thpix.php

Find the average velocity in the following time intervals.

(a) 0 to 4 s
_____ m/s

(b) 0 to 5 s
_____ m/s

(c) 4 s to 5 s
_____ m/s

(d) 5 s to 7 s
______ m/s

(e) 0 to 8 s
______ m/s

Homework Equations



I thought the equation to use would be x(t_2) - x(t_1) / t_2 - t_1

The Attempt at a Solution



After following that equation, I got 5 m/s for (a), but it was incorrect.

Could someone PLEASE tell me what I did wrong. Thanks.
 
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Hi chanv1,

chanv1 said:

Homework Statement



The position versus time for a certain particle moving along the x-axis is shown in the figure below.



Find the average velocity in the following time intervals.

(a) 0 to 4 s
_____ m/s

(b) 0 to 5 s
_____ m/s

(c) 4 s to 5 s
_____ m/s

(d) 5 s to 7 s
______ m/s

(e) 0 to 8 s
______ m/s


Homework Equations



I thought the equation to use would be x(t_2) - x(t_1) / t_2 - t_1

The Attempt at a Solution



After following that equation, I got 5 m/s for (a), but it was incorrect.

Your equation looks right to me; but I am not getting 5m/s for part a. What numbers are you plugging into your equation (for x(t2), x(t1), t2, and t1) to get 5m/s?
 
I did

5(4) - 0(0) / 4-0 = 5

What answer did you come up with? and would you please show me how?
 
chanv1 said:
I did

5(4) - 0(0) / 4-0 = 5

What answer did you come up with? and would you please show me how?

I think you are visualizing the formula incorrectly. It is:

<br /> v_{\rm ave} = \frac{x(t_2)-x(t_1)}{t_2-t_1}<br />
and x(t_2) is not x times t_2. So it might be better to write it as:

<br /> v_{\rm ave} = \frac{x_2-x_1}{t_2-t_1}<br />

because x(t_2) means the position x at time t_2, so x(t_2)=x(4 \mbox{ seconds}) = 5\mbox{ meters}
 
Yeah, I know. I don't understand what I'm doing wrong ... please walk me through this?

Isn't the distance for 4s, 5? so I would 5 * 4 = 20 and so on?

What should I be seeing instead?
 
I seem to be having some computer problems, so I'll repost the edits I made in my last post in case they just are not showing up.

it might be better to write the average velocity formula as:

<br /> v_{\rm ave} = \frac{x_2-x_1}{t_2-t_1}<br />

because x(t_2) means the position x at time t_2, so x(t_2)=x(4 \mbox{ seconds}) = 5\mbox{ meters}, not 20.
 
so would the answer then be 1.25 or 1.3?

but that answer was incorrect too.
 
chanv1 said:
so would the answer then be 1.25 or 1.3?

but that answer was incorrect too.

Unless I'm just not seeing something, it looks like 1.25m/s is the correct answer to me. (Since you mentioned 1.3m/s, did you try 1.25 or did you input the rounded answer?)
 
yeah, I rounded the number. Silly me!
Thanks for all your help!
 
  • #10
Sure, glad to help!
 

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