Calculating Average Velocity from Position vs Time Graph

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The discussion focuses on calculating average velocity from a position versus time graph for a particle moving along the x-axis. Participants clarify the correct application of the average velocity formula, emphasizing that x(t) refers to the position at a specific time rather than multiplying position by time. The initial calculation of 5 m/s for the first interval is identified as incorrect, with the correct interpretation leading to an average velocity of 1.25 m/s. Rounding errors are also noted as a potential source of confusion in the calculations. Accurate understanding of the formula and careful input of values are crucial for obtaining the correct average velocity.
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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!
 
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Sure, glad to help!
 

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