Physics question

[Last One In]

a. how do you find the particles speed at t=10s and t=20s
b. how do you find the distance traveled in the 20 seconds?
Check out the graph;

 

[CyberPitz]

Graph..soo..small..
gah.

 

[Ikerous]

What's that the graph of? Position or velocity?

If its position, then the slope of the graph at 10 will be the speed

 

[Last One In]

Acceleration vs. time. ill repost the graph real quick.

 

[Ikerous]

If its acceleration vs time, then the velocity at 10 is the area between the graph and the x axis from 0 to 10
so at t=10, the velocity is 20 and at t=20 the velocity is 5

 

[CyberPitz]

Man Ikerous, you'd think all the times we make love that I'd get some of that intelligence through Osmosis or something.

 

[Last One In]

If its acceleration vs time, then the velocity at 10 is the area between the graph and the x axis from 0 to 10

What about at the area where the acceleration dips to zero? How do you find the velocity there? wouldn;t it be 20 m/s because the acceleration went to zero at 20 m/s. If you found the area in that case it would say the velocity is zero but it would infact be 20 m/s

 

[Ikerous]

What about at the area where the acceleration dips to zero? How do you find the velocity there? wouldn;t it be 20 m/s because the acceleration went to zero at 20 m/s. If you found the area in that case it would say the velocity is zero but it would infact be 20 m/s

But thats just the change in velocity from 10-15, which is zero
But like you said, before it dips to zero (from 0->10), the velocity gets to 20
so to get the velocity from 0-15 its just 20 + 0

 

[EC]

If I'm reading that graph correctly, then the acceleration from 0 to 10 seconds is 2 m/s^2. Providing I know what I'm talking about, the velocity at 10 seconds is 2m/s^2 * 10s, or 20 m/s. From 10 to 15, the acceleration is 0, so the velocity is constant. Then, from 15 to 20, the acceleration is -3 m/s^2. -3 m/s^2 * 5 s= -15 m/s. The velocity at 20 seconds is 20 m/s - 15 m/s = 5 m/s.

As for the distance, I can't think that much right now. Would anyone like to make sure I did everything correctly for the first part?

 

[Last One In]

Velocity is a vector quantity and cannot be added. But in this case I'm trying to find speed, which is scalar, so it would be correct to add them. Ok, so the speed for the graph at 20s would be (2*10)+0+(-3*5)=5m/s. So do you have any idea of how to find the distance?

 

[Ikerous]

To find the distance, you take the area under between the graph of the velocity, and the x axis from 0 to 20

 

[dfc05]

For distance I'd just use the x = x0 + v0t + .5a*t^2. Except do it 4 times (one for each region of constant acceleration), and carry over x and v each time. But there's probably an easier way.

edit: yeah, do what Ikerous said just graph the velocity

 

[Last One In]

To find the distance, you take the area under between the graph of the velocity, and the x axis from 0 to 20

So I would graph my velocities and find the area of them of a velocity vs. time graph? Isn't there a way to do it with differentiation or something too?

 

[Ikerous]

So I would graph my velocities and find the area of them of a velocity vs. time graph? Isn't there a way to do it with differentiation or something too?

What you're technically doing is integration (Which is just another method for finding the area under stuff)

But it doesnt really work that well here because you dont have the accleration function since the graph isnt exactly a function

 

[Last One In]

What you're technically doing is integration

Is there any chance you could tell me a bit about that or should I just use the graph way?

 

[Ikerous]

Is there any chance you could tell me a bit about that or should I just use the graph way?

If you were given a function for a(t) itd be really easy to do with integration, but since its just random lines, its easiest to use the graph

Or could integrate each line segment and add them, but itd be more tedious that way

 

[Dan]

speed is the 1st integral of acceleration
(assuming you start at a speed of zero)
at t=10 v=20
at t=20 v=10

distance is the 2nd integral of acceleration
(assuming you start at position 0)
d=275

 

[Ikerous]

^ at t=20 v should be 5

 

[Last One In]

I know I'm pressing my luck, but I figure you enjoy this stuff quite a bit. Anyway, i have another problem: A 50.0g bounceball traveling at 25.0m/s bounces off a brick wall and rebounds at 22.0 m/s. If the gball is in contact with the wall for 2.50ms, what is the magnitude of the average acceleration of the ball during this time interval. To solve this I took 25.0 m/s +22. m/s and divided that by 3.5*10^-3s. It gives the correct answer but i'm not really sure if this is the correct method. What is your thought>?

 

[Ikerous]

Seems logical enough to me
(Although you changed the time from 2.5ms to 3.5ms...)

 

[Last One In]

opps, I crapped my pants. Thanks ike, you sure do know your stuff. Maybe you should start your own business or something.

 

[Dan]

yeah sorry, I just looked at the graph without reading the numbers. In that case the distance at t=20 should be 262.5 not 275

 

[Ikerous]

opps, I crapped my pants. Thanks ike, you sure do know your stuff. Maybe you should start your own business or something.

I had a really insane calc1 teacher who made us do all kinds of things with linear motion, so after about a thousand problems on the subject, I should have it down

(I'm prolly going to do awful in my physics 230 class though cuz the teacher is insane ^_^)