3u question( motion , integration) (1 Viewer)

[hkdrmark]

20) A particle moves in a straight line . At time t seconds its displacement is x metres from a fixed point O on the line, its acceleration is a ms^-2, and its velocity is v ms^-1 where v is given by v=32/x - x/2.

a) find an expression for a in terms of x.

b) Show that t = integrate(2x / (64-x^2)), and hence show that x^2 = 64 - 60 e^-t.


I've done question a,

a)
v^2 = 1024/x^2 - 2(32/x * x/2) + x^2/4
v^2 = 1024/x^2 + x^2/4 -32
1/2 v^2 = 512/x^2 + x^2/8 -16
d/dx 1/2 v^2 = -1024/x^3 + x/4
a= x/4 -1024/x^3

b)
v = dx/dt = (64-x^2)/2x
dt/dx = 2x/(64-x^2)
t = integrate (2x/(64-x^2))
t= -ln(64-x^2) +C
C-t = ln(64-x^2)
e^(C-t) = 64-x^2
x^2 = 64 -e^(C-t)

then what should I do to find out the C

 

[Trebla]

You're meant to sub in an initial condition (when t = 0, x = ... to find C). Not quite sure what the initial condition is give here but assuming you typed the question correctly it looks as if it starts initially at x = 2 or x = -2? Or is it meant to be "x^2 = 64 - 64e^-t" where the particle is initially at the origin?

 

[namburger]

Insufficient information.
If they gave actual values like where the particle was initially, than it will be possible. Maybe the answer is expected to be in pronumerals?
Im not sure

 

[YannY]

Insufficient information.
If they gave actual values like where the particle was initially, than it will be possible. Maybe the answer is expected to be in pronumerals?
Im not sure

shit, nam too smart - done this shit already

 

[namburger]

shit, nam too smart - done this shit already

Nah im too stupid - you're up to 4u integration zz

 

[YannY]

how do you know ive done 4u integration? unless youve already done it. haha