Position of particle two dimensions

[HATLEY1997]

For part c, I have found that 0.0175i is added on per second however the j-value only changes slightly and is different each time. Is there somewhere I am going wrong or am I okay to round it?

 

[mario99]

For part c, I have found that 0.0175i is added on per second however the j-value only changes slightly and is different each time. Is there somewhere I am going wrong or am I okay to round it?

For Part C, you have two points:

[imath](1,3)[/imath] and [imath](1+\sin t,3-\cos t)[/imath]

The first point is fixed while the second point is changing, but the distance between them is always [imath]1[/imath].

Proof: use the distance formula.

[imath]d = \sqrt{(1+\sin t - 1)^2 + (3-\cos t - 3)^2} = \sqrt{(\sin t)^2 + (-\cos t)^2} = \sqrt{1} = 1[/imath]

It does not matter what is the time. Let us take a random time, [imath]t = 5[/imath].

[imath]d = \sqrt{(1+\sin 5 - 1)^2 + (3-\cos 5 - 3)^2} = \sqrt{(\sin 5)^2 + (-\cos 5)^2} = \sqrt{1} = 1[/imath]

 

[HATLEY1997]

For Part C, you have two points:

[imath](1,3)[/imath] and [imath](1+\sin t,3-\cos t)[/imath]

The first point is fixed while the second point is changing, but the distance between them is always [imath]1[/imath].

Proof: use the distance formula.

[imath]d = \sqrt{(1+\sin t - 1)^2 + (3-\cos t - 3)^2} = \sqrt{(\sin t)^2 + (-\cos t)^2} = \sqrt{1} = 1[/imath]

It does not matter what is the time. Let us take a random time, [imath]t = 5[/imath].

[imath]d = \sqrt{(1+\sin 5 - 1)^2 + (3-\cos 5 - 3)^2} = \sqrt{(\sin 5)^2 + (-\cos 5)^2} = \sqrt{1} = 1[/imath]

Thank you

 

[Dr.Peterson]

For part c, I have found that 0.0175i is added on per second however the j-value only changes slightly and is different each time. Is there somewhere I am going wrong or am I okay to round it?

Although this is irrelevant to the question itself, I notice that you are taking sines and cosines in degree mode, rather than radians, which is assumed for a problem like this. [imath]\sin(1rad) = 0.84147\dots[/imath]; [imath]\sin(1^\circ) = 0.01745...\dots[/imath]

 

[HATLEY1997]

Although this is irrelevant to the question itself, I notice that you are taking sines and cosines in degree mode, rather than radians, which is assumed for a problem like this. [imath]\sin(1rad) = 0.84147\dots[/imath]; [imath]\sin(1^\circ) = 0.01745...\dots[/imath]

View attachment 38478

That’s really helpful and something I forgot to change on my calculator. Thank you

 

[mario99]

Thank you

You are welcome.