please helpp :( (1 Viewer)

posh bitch · Oct 8, 2012

some one please help me answer part (b) of this question
i have no clue !

see attached doc.

Come at me · Oct 8, 2012

sub in x=0 and pi/4 and you should not be able to work it out

therefore ......

Sy123 · Oct 8, 2012

posh bitch said:
some one please help me answer part (b) of this question
i have no clue !

see attached doc.

One way to do it is (might not be very elegant but..)

$v=\frac{5}{\sin 2x} \\ \frac{dt}{dx}=\frac{1}{5}\sin 2x \\ \\ t=\int 0.2 \sin 2x \ dx \\ t=\frac{-1}{10}\cos 2x$

For all real time t, t > 0

$-0.1 \cos \2x > 0 \\ \therefore \cos 2x <0 \\ \\ \therefore $For all x$ \ \ 0<x<\pi / 4 \\ t<0 \therefore $particle cannot travel between x=0 and Pi over 4$$

Come at me said:
sub in x=0 and pi/4 and you should not be able to work it out

therefore ......

This is incorrect

please helpp :( (1 Viewer)

posh bitch

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Come at me

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Sy123

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