Originally Posted by

**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