# Thread: Question about integration + motion

1. ## Question about integration + motion

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

2. 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
$x^2 = 64 - e^{C-t} = 64 - e^{C} e^{-t} = 64 - A e^{-t}$.

To get A you need a boundary condition that gives the value of x for a value of t. An obvious one to consider is x = 0 when t = 0, but v (and a) is undefined at x = 0 so that's no good .....

The answer you have to get seems to assume the boundary condition x = 2 when t = 0. From the information given there's no good reason where this boundary condition has come from ....