Kinematics of Particles: What is the Particle's Acceleration at 68 s?

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SUMMARY

The discussion focuses on calculating the particle's acceleration and velocity at 68 seconds based on its position equations: x = -34t^4 - 28t^3 + 7 and y = 25t^2 + 13t^3 + 5. The correct velocity at 68 seconds is determined to be v_p = (-43200000 m) i + (184000 m) j. To find acceleration, participants clarify that it is calculated using the second derivatives of the position functions, specifically a(t) = (d²x/dt²)i + (d²y/dt²)j.

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roam
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Homework Statement



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

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

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

(736000000m) i + (4200000m)j

(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

v_p = (43200000m) i + (184000m)j

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

\Delta t = t_f-t_i=68-0 =68

\Delta x = x_f-x_i = 736000000 - 7 = 375999993

So, \frac{375999993}{68}=10823529.31.

This is not the right answer for the i component. What is the problem?
 
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just use v(t)= (dx/dt)i+(dy/dt)j
 
rock.freak667 said:
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...
 
Can you show exactly how you did it? You may have made a mistake in the algebra somewhere.
 
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 (736000000m) i + (4200000m)j

\Delta t = t_f-t_i=68-0 =68

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

\Delta x = x_f-x_i = 736000000 - 7 = 375999993

So, v=\frac{\Delta x}{\Delta t}=\frac{375999993}{68}=10823529.31.
 
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?
 
rock.freak667 said:
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?

\frac{375999993}{68}=10823529.31

If this not true then I have no idea...
 
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?
 
rock.freak667 said:
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
roam said:
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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