Intermediate dynamics problems

[jazzyfizzle]

1. A

 

[Doc Al]

Find the acceleration, then use a bit of kinematics.

 

[Doc Al]

But that's the thing.. I don't know the force. How are you supposed to even do this when there's no numbers involved ?

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

I just tried it and got :

x=2[(1/2)(F0/m)t^2]

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.

 

[Doc Al]

Should I maybe use vf^2=vi^2+2ax to find the final velocity of the first segment?
then go from there ?

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

 

[jazzyfizzle]

For a final answer I got

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

But... I don't think that's right at all...

 

[Doc Al]

For a final answer I got

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

But... I don't think that's right at all...

No, it's not.

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

 

[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)/mthen I used that final velocity as the initial velocity for the 2nd segment
and plugged in a=2F0/m
into x=v0t+1/2at^2and somehow came out with the answer
x=((2F0*t0^2)/m)) + (4F0t0)/mWhere did I go wrong ?

 

[jazzyfizzle]

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

 

[Doc Al]

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

Good! You have the first segment done.

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

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

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

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

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

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

 

[jazzyfizzle]

Ooh, well I never went back and added the first segment.. maybe that's 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.

 

[Doc Al]

Redo and then simplify your results for the second segment.

 

[jazzyfizzle]

For the second segment I'm getting the distance as

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

 

[Doc Al]

For the second segment I'm getting the distance as

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

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

 

[jazzyfizzle]

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

 

[Doc Al]

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

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

 

[jazzyfizzle]

Oooh. That makes sense. Duh.