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Physical Application of Cal (1 Viewer)

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once you get the expression for velocity
change cos2t into 2cos^2t - 1

then you have v=cost-2cos^2t-1
set v=0
and the discriminant is less than zero, therefor never = 0

therefore never at rest
 
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lyounamu

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tacogym27101990 said:
once you get the expression for velocity
change cos2t into 2cos^2t - 1

then you have v=cost-2cos^2t-1
set v=0
and the discriminant is less than zero, therefor never = 0

therefore never at rest
That's brilliant. Thanks.

By the way, what about this one:

A body is moving with simple harmonic motion in a straight line. It has an amplitude of 10 metres
and a period of 10 seconds. How long would it take for the body to travel from one of the extremities of its path of motion to a point 4 metres away?

 

Aerath

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Oh, changing cos2t using double angle formula was genius.
 

lyounamu

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Aerath said:
Oh, changing cos2t using double angle formula was genius.
Yeah, that's something to note. Really good question, that was.
 

vds700

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lyounamu said:
That's brilliant. Thanks.

By the way, what about this one:

A body is moving with simple harmonic motion in a straight line. It has an amplitude of 10 metres
and a period of 10 seconds. How long would it take for the body to travel from one of the extremities of its path of motion to a point 4 metres away?

wow someones pretty busy lol.

let x = 10sin(pi/5)t
Sketch this graph
when x = 10 (extremity)
10 = 10sin(pi/5)t
sin(pi/5)t = 1
(pi/5)t = pi/2
t = 5/2

when it is 4m away from this point, it is 6m away from the origin.
6 =10sin(pi/5)t
sin(pi/5)t = 3/5
(pi/5)t = 0.6435, 2.4981
t = ..., 3.9758 we are only interested in the second solution as this is when it is at 6m after being at 10m.

Time traken to be 4m away = 3.9758 - (5/2) = 1.4758 s.
 

lyounamu

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vds700 said:
wow someones pretty busy lol.

let x = 10sin(pi/5)t
Sketch this graph
when x = 10 (extremity)
10 = 10sin(pi/5)t
sin(pi/5)t = 1
(pi/5)t = pi/2
t = 5/2

when it is 4m away from this point, it is 6m away from the origin.
6 =10sin(pi/5)t
sin(pi/5)t = 3/5
(pi/5)t = 0.6435, 2.4981
t = ..., 3.9758 we are only interested in the second solution as this is when it is at 6m after being at 10m.

Time traken to be 4m away = 3.9758 - (5/2) = 1.4758 s.
Thanks, Andrew. At this rate, I will get my ass kicked in Trials by everyone in 4 Unit class. I don't any miracle like 100% in MX1 will ever happen again. I am gone....
 

vds700

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lyounamu said:
Thanks, Andrew. At this rate, I will get my ass kicked in Trials by everyone in 4 Unit class. I don't any miracle like 100% in MX1 will ever happen again. I am gone....
no worries man. Dw you'll be fine, i have barely done much study for MX1, and catholic paper has the potential to be an absolute nightmare lol. I'm not expecting anywhere near 100%.
 

Jachie

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breaking said:
more like physical appreciation of cal.
more like jachie.
ahahahahaha fucking lol
 

hon1hon2hon3

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Just to make things clear"Er" , the period is 10 , soo T = 10 . T also equals to (2 pie)/n , thus n = pie/5 . Which is what is used in that solution above
 

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