Calculating Average Velocity from Position vs Time Graph

[chanv1]

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.

 

[alphysicist]

Hi chanv1,

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?

 

[chanv1]

I did

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

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

 

[alphysicist]

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:

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

$$v_{\rm ave} = \frac{x_2-x_1}{t_2-t_1}$$

because $x(t_2)$ means the position x at time $t_2$, so $x(t_2)=x(4 \mbox{ seconds}) = 5\mbox{ meters}$

 

[chanv1]

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?

 

[alphysicist]

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:

$$v_{\rm ave} = \frac{x_2-x_1}{t_2-t_1}$$

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.

 

[chanv1]

so would the answer then be 1.25 or 1.3?

but that answer was incorrect too.

 

[alphysicist]

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?)

 

[chanv1]

yeah, I rounded the number. Silly me!
Thanks for all your help!

 

[alphysicist]

Sure, glad to help!