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

AI Thread Summary
The discussion focuses on calculating the velocity and acceleration of a particle moving in a plane, with its position defined by specific equations. At t = 68 seconds, the particle's displacement is given, but there is confusion regarding the correct calculation of velocity, particularly the i component, which should be negative. Participants clarify that to find velocity, one must differentiate the position equations, leading to the correct expressions for dx/dt and dy/dt. For acceleration, the same differentiation method applies, using the second derivatives of the position equations. The conversation emphasizes the importance of careful algebraic manipulation and differentiation in solving kinematic problems.
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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