1. Aug 30, 2004

### Shay10825

Hi

Find the acceleration.

from x= 2m to x= 8m during a 2.5-s time interval.

I found the slope to be 2.8 so the velocity is 2.8. Now how do I fond the acceleration? I know you have to find the slope of the velocity to get the acceleration, but i keep geeting the wrong answer .

~Thanks

2. Aug 30, 2004

### Tide

I think we're missing some information or you provided incorrect numbers.

If the acceleration is 0.32 m/s^2 then an object will move 1 m in 2.5 seconds under uniform acceleration and starting from rest.

3. Aug 30, 2004

### Shay10825

the problem said:

A particle confined to motion along the x axis moves with constant acceleration from x= 2m to x= 8m during a 2.5-s time interval. The velocity of the partical at x=8 m is 2.8 m/s. What is the acceleration during this time interval?

I know how to do it with the kinematic equations but I want to do it without them. How would you do it with derivatives?

4. Aug 30, 2004

### HallsofIvy

Staff Emeritus
You left the "The velocity of the partical at x=8 m is 2.8 m/s." out of the original post- that's the "missing information" Tide was talking about.

If we take the constant acceleration to be "a", and initial velocity to be "v0", then the accleration after t seconds is at+ v0 and the postion is
(1/2)at2+ v0t+ 2 (the original position was x= 2).

You are told that when t= 2.5, (1/2)a(2.52)+ v0(2.5)+ 2= 8 and a(2.5)+ v0= 2.8.

Solve those two equations for a and v0.
at+ v0[/b

5. Aug 30, 2004

### Shay10825

Sorry .
How come when you find the slope of the v= 2.8 and t=2.5 it does not equal the acceleration of .32 ?

6. Aug 30, 2004

### needhelpperson

because you need another point to determine the slope, not just when v = 2.8. and t = 2.5. You can determine the initial velocity and use that as the second point.

7. Aug 30, 2004

### Shay10825

so (Vf-Vi)/2.5 = acceleration?

but how come (2-8)/2.5 does not = 2.8? How would I fond the slope for the velocity?

8. Aug 30, 2004

### needhelpperson

because there's an acceleration, the slope for velocity is always changing, thus, you can find the average velocity using (y2-y1)/(x2-x1).

-->(8-2)/(2.5-0) = average velocity.

9. Aug 30, 2004

### Shay10825

How would I find the slope or derivative of the velocity at 2.8m/s and t=2.5 (the answer is .32m/s^2 but how would we find this ).

Last edited: Aug 30, 2004
10. Aug 30, 2004

### needhelpperson

no need for calculus here. Besides, they already told you the slope for the velocity at t = 2.5. It's 2.8m/s. But that's not what you're asking for, since .32m/s^2 is the acceleration.

figure out vi using (vi+vf)/2 = average velocity.

You already know the average velocity.

once you have it. a = (vf-v1)/(2.5)

11. Aug 30, 2004

### Shay10825

I understand all of that but say if this was a different problem and they did not give you the 2.8. How would you find it (2.8) using derivitaves and the information (2 m, 8 m and 2.5 sec)?

Last edited: Aug 30, 2004
12. Aug 30, 2004

### needhelpperson

Then you wouldn't be able to solve it. There's not enough information given.

13. Aug 30, 2004

### Shay10825

so if you had A particle confined to motion along the x axis moves with constant acceleration from x= 2m to x= 8m during a 2.5-s time interval. Find the velocity of the partical at x=8 m you would not be able to solve it because there is not enough info?

14. Aug 30, 2004

### needhelpperson

that's correct.

15. Aug 30, 2004

### Shay10825

But I thought the slope or derivative of the displacement or position time graph is the velocity at that time? If it did not say "a 2.5 s time interval" but said specific points then would you be able to find the derivative or the velocity?

Last edited: Aug 30, 2004
16. Aug 30, 2004

### needhelpperson

Yes, then can you figure it out... if you're given a graph and all...

17. Aug 30, 2004

### Shay10825

THANK YOU SOOOOOO MUCH!!!!!! I just started AP Physics B and it's all new and so hard .

Last edited: Aug 30, 2004