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Mathematics: Post your doubts here!

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Nov 2012 .

Ah all right, so it's asking for the speed of the particle at O, which is where it starts to move from.
Obviously at this point the displacement will be 0, so we have to integrate the equation given for velocity and take it as 0.
This is because of the rule, which I'm assuming you know:

Displacement ---> Velocity ---> Acceleration (Differentiate)
Displacement <--- Velocity <--- Acceleration (Integrate)

Once you integrate it and take it as 0, you should get t = 80.
Now just put this value of t back into your equation for velocity, it should give you your speed.
Hope you got it x)
 
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Ah all right, so it's asking for the speed of the particle at O, which is where it starts to move from.
Obviously at this point the displacement will be 0, so we have to integrate the equation given for velocity and take it as 0.
This is because of the rule, which I'm assuming you know:

Displacement ---> Velocity ---> Acceleration (Differentiate)
Displacement <--- Velocity <--- Acceleration (Integrate)

Once you integrate it and take it as 0, you should get t = 80.
Now just put this value of t back into your equation for velocity, it should give you your speed.
Hope you got it x)
i got it Alhamdulelah , A big thanks :)
 
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what paper do u recommend it to me to do it , as i don't have enough time .. a paper has a lot of ideas ?
 
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It's given you the maximum velocity, at max velocity acceleration is 0.

So, since we have the equation for velocity , we need to change it to acceleration by differentiating it.
This gives us , 12ot-3t^2
We take this as 0 and find the value of t

120t-3t^2 = 0
3t(40t-4) = 0
t= 40

Now that we have a value of t , replace it in the original equation for velocity and take velocity as 6.4 since thats it's max.
k(60t^2-t^3) = 6.4
k(32000) = 6.4
k= 6.4/32000
k= 0.0002

Hope you get it x)

i got it Alhamdulelah , A big thanks :)

You're welcome.
 
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yeah cause i screwed up bad with pure and really hope mechanics gets me the A .... so no not at all :p
 
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