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

[roam]

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?

 

[rock.freak667]

just use v(t)= (dx/dt)i+(dy/dt)j

 

[roam]

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...

 

[rock.freak667]

Can you show exactly how you did it? You may have made a mistake in the algebra somewhere.

 

[roam]

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, $$v=\frac{\Delta x}{\Delta t}=\frac{375999993}{68}=10823529.31$$.

 

[rock.freak667]

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?

 

[roam]

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.

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...

 

[rock.freak667]

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?

 

[roam]

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?

 

[rock.freak667]

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