Calculating Displacement from a Velocity-Time Graph

[oldspice1212]

Hey guys, so I'm having some trouble, not sure if back of the book is wrong or if I'm doing it wrong, probably me though lol.

Alright so I have this graph,

http://imageshack.us/photo/my-images/27/wqqf.jpg/It's asking me to find displacement at 10.0s.

So I know that displacement = area of a velocity- time graph,

therefore at first I assumed it was a trapezoid but that didn't make sense so I used area formulas separately and am getting,

1/2bh+bh+1/2bh

Using this formula I get 39 m/s² East but in the back it shows as 36 m/s² East.1/2(5)(6)+(3)(6)+1/2(2)(6)

 

[Simon Bridge]

You need to get a bit more disciplined in your use of formulas ... 1/2bh+bh+1/2bh
(= A) makes no sense since it basically says that the total area is 2bh.

You want to say something like:
$\small d=\frac{1}{2}vT_1 + vT_2 + \frac{1}{2}vT_3$

The idea is to understand the physics, then use the equations in such a way that you are communicating what you are doing... that way you'll have more confidence in your results.

To see where you are going wrong, or right for that matter, you need to show your working.

 

[Dick]

Hey guys, so I'm having some trouble, not sure if back of the book is wrong or if I'm doing it wrong, probably me though lol.

Alright so I have this graph,

http://imageshack.us/photo/my-images/27/wqqf.jpg/


It's asking me to find displacement at 10.0s.

So I know that displacement = area of a velocity- time graph,

therefore at first I assumed it was a trapezoid but that didn't make sense so I used area formulas separately and am getting,

1/2bh+bh+1/2bh

Using this formula I get 39 m/s² East but in the back it shows as 36 m/s² East.


1/2(5)(6)+(3)(6)+1/2(2)(6)

Yes, you are right. Except the units are m. Not m/s^2.

 

[oldspice1212]

Yes, you are right. Except the units are m. Not m/s^2.

Thought so, and yes m, lol sorry about that, been doing all these acceleration questions must have put it by habbit.


@simon Bridge, I actually did do it that way, but was too lazy to put it like that on the forums ;).

 

[siddharth23]

A better way to do it, for understanding, is divide it into variable velocity and constant velocity parts, and calculate the displacement separately. Just saying area is the displacement doesn't with understanding.

 

[Simon Bridge]

@simon Bridge, I actually did do it that way, but was too lazy to put it like that on the forums ;).

Oh OK - then you should have had more confidence in your result.
The underlying question here is, "How can you tell when you have the right answer - without having to use some authority?"

 

[oldspice1212]

Oh OK - then you should have had more confidence in your result.
The underlying question here is, "How can you tell when you have the right answer - without having to use some authority?"

You're right, I thought I was right for quite some time, but other times I felt uncertain since when the book has all the right answers and you look and it's wrong. It also doesn't hurt to double check on the forums, right? :)

 

[Simon Bridge]

There's also the metadata - 36 vs 39, especially if the book is old enough to have been set in movable type ... someone just read the number upside down: it's a plausible typo.

Interestingly it can hurt ...if the responses are along the lines of "yeah you got it right".
Asking on forums can become a crutch - take care huh?

While you have this resource - you can make full use of it - post the complete working, lay it out, show all your reasoning, look for ways to improve how you go about things. There is so much we do here.