# Intermediate Dynamics - HELP!

1. Sep 7, 2010

### jazzyfizzle

1. A

Last edited: Sep 7, 2010
2. Sep 7, 2010

### Staff: Mentor

Find the acceleration, then use a bit of kinematics.

3. Sep 7, 2010

### Staff: Mentor

Since you don't have numbers, you'll express your answer in terms of the given data--F0, t0, and m.

You're on the right track. Hint: Treat each segment of the motion separately. In the first segment (from t = 0 to t0), the initial speed is zero--but that's not the case for the second half of the motion.

Hint2: Consider the average speed during each segment.

4. Sep 7, 2010

### Staff: Mentor

There's a much easier way to find the final velocity of the first segment. What's the definition of acceleration?

5. Sep 7, 2010

### jazzyfizzle

For a final answer I got

x=((2F0*t0^2)/m)) + (4F0t0)/m

But... I don't think thats right at all...

6. Sep 7, 2010

### Staff: Mentor

No, it's not.

Do it step by step, one segment at a time.

7. Sep 7, 2010

### jazzyfizzle

Ok what I did was :

First I used x=v0t+1/2at^2 to find the distance traveled for the first segment before the force doubled

I plugged in a=(F/m) for the acceleration and v0=0 for the initial velocity
Then i got
x=1/2(F0/m)t^2
for the first segment

Then,
i used vf^2=vi^2+2ax to find the final velocity of the first segment
(plugging in the x i found from above and a=F/m and v0=0)
which i got to be
vf= (F0t)/m

then I used that final velocity as the initial velocity for the 2nd segment
and plugged in a=2F0/m
into x=v0t+1/2at^2

and somehow came out with the answer
x=((2F0*t0^2)/m)) + (4F0t0)/m

Where did I go wrong ?

8. Sep 7, 2010

### jazzyfizzle

The definition of acceleration is the change in velocity over the change in time

9. Sep 7, 2010

### Staff: Mentor

Good! You have the first segment done.

Good. An easier way would be to use vf = vi + at, but your way is perfectly fine.

That will give you the distance for the second segment. (Then you'll add that to what you already found for the first segment.)

I don't see how you got this. Show me what you plugged in where.

10. Sep 7, 2010

### jazzyfizzle

Ooh, well I never went back and added the first segment.. maybe thats where I went wrong ? .. and my algebra could be a little off too. I'll probably need to go back and double check that also.

11. Sep 7, 2010

### Staff: Mentor

Redo and then simplify your results for the second segment.

12. Sep 7, 2010

### jazzyfizzle

For the second segment I'm getting the distance as

[F0(3t0 + t0^2)]/m

13. Sep 7, 2010

### Staff: Mentor

That's not dimensionally correct. Show me what you plugged in for each term in the following:
x=v0t+1/2at^2
x = (v0)(t) + 1/2(a)(t)^2

14. Sep 7, 2010

### jazzyfizzle

v0 ---> (F0t0)/m
t ----> 2
a -----> (2F0)/m

15. Sep 7, 2010

### Staff: Mentor

v0 and a are correct, but the time for the second segment is the same as the first: t0. (From t0 to 2t0.)

16. Sep 7, 2010

### jazzyfizzle

Oooh. That makes sense. Duh.