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Kinematics of Particles

  • Thread starter roam
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  • #1
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Homework Statement



A particle moves in a plane. The particle (x,y) position on the plane is given by:

[tex]x = -34t^4 − 28t^3 + 7[/tex]
[tex]y = 25t^2 + 13t^3 + 5[/tex]

Therefore the particle's displacement [from the (x,y) origin) at time t = 68 s is

[tex](736000000m) i + (4200000m)j[/tex]

(a) What is the particle's velocity at 68 s ?

(b) What is the particle's acceleration at 68 s ?

The Attempt at a Solution



(a) for this part the correct answer has to be

[tex]v_p = (43200000m) i + (184000m)j[/tex]

But I can't see how they have arrived at this answer! The velocity is [tex]\frac{\Delta x}{\Delta t}[/tex]

[tex]\Delta t = t_f-t_i=68-0 =68[/tex]

[tex]\Delta x = x_f-x_i = 736000000 - 7 = 375999993[/tex]

So, [tex]\frac{375999993}{68}=10823529.31[/tex].

This is not the right answer for the i component. What is the problem?
 

Answers and Replies

  • #2
rock.freak667
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just use v(t)= (dx/dt)i+(dy/dt)j
 
  • #3
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just use v(t)= (dx/dt)i+(dy/dt)j
That's exactly what I did! And I don't know why I get the wrong answer...
 
  • #4
rock.freak667
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Can you show exactly how you did it? You may have made a mistake in the algebra somewhere.
 
  • #5
1,266
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Can you show exactly how you did it? You may have made a mistake in the algebra somewhere.
Here is my working for the i component:

The particle's displacement is [tex](736000000m) i + (4200000m)j[/tex]

[tex]\Delta t = t_f-t_i=68-0 =68[/tex]

Since for t=0 the first equation, [tex]x=-34t^4-28t^3+7[/tex], will be 7.

[tex]\Delta x = x_f-x_i = 736000000 - 7 = 375999993[/tex]

So, [tex]v=\frac{\Delta x}{\Delta t}=\frac{375999993}{68}=10823529.31[/tex].
 
  • #6
rock.freak667
Homework Helper
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In your answers are you working with a specified degree of accuracy?

Also if x=-34t4-28t3+7, how is your i component positive?

Alos what is dx/dt and dy/dt equal to in terms of t?
 
  • #7
1,266
11
In your answers are you working with a specified degree of accuracy?
I think 2% is the accuracy tolerance.

Also if x=-34t4-28t3+7, how is your i component positive?
oops, I'm sorry, the answer has to be -43200000 i. :redface:

Alos what is dx/dt and dy/dt equal to in terms of t?
[tex]\frac{375999993}{68}=10823529.31[/tex]

If this not true then I have no idea...
 
  • #8
rock.freak667
Homework Helper
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If x=t5, then dx/dt is 5t4


So if x=-34t4-28t3+7 and y =25t2+13t3+5

what is dx/dt and dy/dt equal to?
 
  • #9
1,266
11
If x=t5, then dx/dt is 5t4


So if x=-34t4-28t3+7 and y =25t2+13t3+5

what is dx/dt and dy/dt equal to?
Thanks a lot it worked! :)

Finally, the question asks "what is the particle's acceleration at 68 s?". If I divide the particle's velocity at 68 s by time=68 seconds, it doesn't give me the correct answer. So what formula do I need to use?
 
  • #10
rock.freak667
Homework Helper
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31
Thanks a lot it worked! :)

Finally, the question asks "what is the particle's acceleration at 68 s?". If I divide the particle's velocity at 68 s by time=68 seconds, it doesn't give me the correct answer. So what formula do I need to use?
You use the same method.


a(t)=(d2x/dt2)i+(d2y/dt2}j
 

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