Vector notation and force and force Help- test this morning

  • #1
~christina~
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[SOLVED] vector notation and force and force Help- test this morning..

Homework Statement


A time dependent force F(t)= 20t N/s acts along the possitive y axis on a 4.00kg object. The object starts at the origin with the initial velocity

[tex]\vec{}v(0)= (1.00m/s)) \hat{}i-(1.00m/s) \hat{}j[/tex].

a) find the velocity of the object as a function of time.
b) find the position of the object as a function of time
c) what is the speed of the object at time t?

Homework Equations


?? the ones given?



The Attempt at a Solution



I have no idea where to start..it seems simple enough..


can someone please please please help me out?
by this morning to be exact..(I have a test)
 

Answers and Replies

  • #2
Doc Al
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You have the force and the mass, thus you can figure out the acceleration. Given the acceleration as a function of time, how would you find the velocity as a function of time? And then the position? (Hint: Only one component is accelerated. Hint2: A bit of calculus is required.)
 
  • #3
~christina~
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Hm..
so ..acceleration...

F(t)= 20 t N/s
a= ?
m= 4.0kg

F= ma (not sure how t would get into that unless I plug in 20 t N/s... (I think the

notation is confusing me)

(20t N/s)/ 4.0kg= a

a= 5 t m/s ?? (not sure about the notation once again...)

I guess if this is correct that I would just do integration to find the v and then the x

however why was the initial velocity given in the form of vector notation ?

I was thinking of using that....

thanks
 
  • #4
Doc Al
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Hm..
so ..acceleration...

F(t)= 20 t N/s
a= ?
m= 4.0kg

F= ma (not sure how t would get into that unless I plug in 20 t N/s... (I think the

notation is confusing me)

(20t N/s)/ 4.0kg= a

a= 5 t m/s ?? (not sure about the notation once again...)
The acceleration is 5t m/s^2. But in what direction? They tell you: "acts along the positive y axis". That means the acceleration should be written as [itex]5 t \hat{j}[/itex] m/s^2.

I guess if this is correct that I would just do integration to find the v and then the x
Yes.

however why was the initial velocity given in the form of vector notation ?
Because direction matters!
 
  • #5
~christina~
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Because direction matters!

so I don't actually use that equation?

anyways...

a(t)= 5tj m/s^2

v(t)= 2.5tj^2 + t

x(t)= .83 tj^3 + 1/2 t^2 + t

Is this how it's supposed to look?

(Once again the units are getting to me...couldn't figure out the units for the v but technically it should be m/s and x should be in m)

Am I supposed to have anymore j's in the equations??
 
  • #6
Doc Al
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Seems like the j notation is messing you up a bit.
a(t)= 5tj m/s^2
a(t)y = 5t (direction: +y axis (a.k.a j); units: m/s^2)
v(t)= 2.5tj^2 + t
Integrate a(t) once to get v(t):
v(t)y = 2.5t^2 + C (y-component of velocity)

Use the given initial velocity to find the integration constant C:
v(t)y = 2.5t^2 - 1 (y-component only)

The complete v(t) is:
v(t) = (1) i + (2.5t^2 -1) j


x(t)= .83 tj^3 + 1/2 t^2 + t
Integrate v(t) to get the y-component of position:
y(t) = (2.5/3)t^3 -t + C

Use the given info to find the integration constant:
y(t) = (2.5/3)t^3 -t

The x-axis motion is just constant speed:
x(t) = (1)t = t

In vector notation, the position as a function of time would be:
(t) i + ((2.5/3)t^3 -t) j
 
  • #7
~christina~
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good grief...I wasn't even close... it was more complicated than I thought it would be

for part c where I have to find the speed of the object at time t

wouldn't I take the

v(t) = (1) i + (2.5t^2 -1) j

and then find the V from the x and y component's i and j ?

rad ( 1^2 + (2.5t^2 -1)^2) = ....

technically that's what I would think but that looks well...complicated...

Is that correct?

Thanks alot
 
  • #8
Doc Al
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Yes, that's correct.
 
  • #9
~christina~
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Thank you Doc Al
 

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