# Force from differentiation

1. Oct 23, 2007

### ~christina~

1. The problem statement, all variables and given/known data
An object with mass m moves along the x axis. It's position as a function of time is given by

X(t)= At- Bt^2 + Ct^3

A,B,C are constants

a) find expression for the net force acting on the object as a function of time.

b) let A= 2m/s, B= 10m/s^2, and C= 1m/s^3

c) at what position is the net force 0?

d) Sketch graphs of the velocity and force vs time...

2. Relevant equations

F=ma

The equation given
3. The attempt at a solution

a) eqzn for the net force acting on the object as a function of time...

I have not a clue...is it I find the acceleration equation and then go and say m= constant so it doesn't matter?

x(t)=Ct^3 -Bt^2 + At

v(t)= x'(t)= 3Ct^2-2Bt +A

a(t)= v'(t)= 6Ct- 2B ==> is this it but how do I add m??

I assume since m is a constant so only thing changing is a, Am I correct?

b) when net force = 0
plugged in the numbers

x(t)= t^3 - 10t^2 + 2t

a(t)= 6Ct- 2B

a(t) 6t- 20
when is the force 0?? isn't it when accleration= 0?? and time is..

0= 6t-20
20/6= t

t= 3.33s

c) value of velocity when net force = 0

v(t)= x'(t)= 3Ct^2-2Bt +A

v(t)= 3t^2 - 20t + 2

t= 3.33 when net force = 0

plugging in..

v(3.33)= 3(3.33)^2- 20(3.33) + 2

v= -31 m/s

Is this alright???

2. Oct 23, 2007

### mjsd

part a) F=ma so multiply m with the expression of a(t) to get force as a function of t
the rest seems ok

3. Oct 23, 2007

### ~christina~

what?
how would I multiply that into the a??
I just can't get my mind around to it...lets see

F= ma
a(t)= v'(t)= 6Ct- 2B

F(t)= m(6t- 20)

F(t)= 6tm- 20m ???

How can this part be wrong but the other parts okay?
wouldn't I get points taken off if the equation I was using didn't include the m for mass?
I would think so..

4. Oct 23, 2007

### mjsd

you are confusing me with your question(s)....
You were asked to find F(t) given m and x(t) so as u have done, calculate a(t) then
F(t)=m a(t)
in part b) m cancels out anyway since it doesn't depend on t and so won't be affected by whether m is there or not, likewise for the subsequent parts.. except may be the graphing bit which you have not shown here anyway.

5. Oct 23, 2007

### Staff: Mentor

That's all there is to it.

Because it didn't ask for the force, but position and time where force was zero, which is where the acceleration is zero.

In part c, weren't you supposed to find position?

6. Oct 23, 2007

### ~christina~

I forgot to find position...

well since t= 3.33 s is when the force = 0

x(t)= t^3 - 10t^2 + 2t

so

x(3.33)= 3.33^3 - 10(3.33)^2 + 2(3.33)= -67.3m

but wait..why is distance negative...I calculated it twice..

how would the m cancel out in the
F(t)= 6tm - 20m??

when you plug in t would'nt you still have ..m?

Thank You

7. Oct 23, 2007

### stewartcs

The distance is not negative, its position is negative.

The distance travelled will be positive even though the position is negative at the moment the acceleration is zero.

Don't confuse the total distance travelled (the displacement) with an objects position.

8. Oct 23, 2007

### Staff: Mentor

distance vs displacement

Just to nitpick on terminology a bit: I would say that the displacement (which is -67.3m, according to her calculations) does give you the final position (with respect to the starting point). Displacement is a vector describing change in position, whereas distance is a scalar representing the cumulative "ground" you've covered.

But your point remains. While "distance" traveled cannot be negative, the position (and displacement) certainly can.

9. Oct 23, 2007

### stewartcs

I would have to say that you are indeed correct Doc Al, the displacement may certainly be negaitve. I should not have used displacement there, but rather distance.

Thanks for pointing that out.

10. Oct 23, 2007

### ~christina~

So the distance = 67.3m
but the displacement= -67.3m

Is this..what you people were saying?
If it is then I get it.

11. Oct 23, 2007

### mdk31

Not necessarily. For example, if an object moves from position 0 m to position 300 m, and from position 300 m to a final position -290 m, the total displacement is -290 but the total distance traveled is certainly not +290. The total distance traveled would be 890

Last edited: Oct 23, 2007
12. Oct 23, 2007

### Staff: Mentor

13. Oct 23, 2007

### ~christina~

so the displacement is just d= -67.3m but I don't know the distance? or do I?

Hm..it explained vectors and scalars and displacement and distance however if it was just an equation like this problem was..I don't exactly have a picture of what is happening...

Last edited: Oct 23, 2007
14. Oct 23, 2007

### Staff: Mentor

Since you have the position as a function of time, you can figure out the distance traveled between any two times. You'd have to examine the path it takes. (Make your own picture.) For example, at T = 0, the particle is at X = 0 and has a velocity of 2 m/s. Figure out how far it gets before it turns around, and just add up all the distance segments between each turn around point.

Not as simple as finding the final position, but you can do it.

15. Oct 23, 2007

### ~christina~

um..I don't think I need to find it for this problem so I'll work on my other ones first..however I ever need to find the distance traveled I know just what to do.

Thanks for explaining that