Question #0ca28

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Answer 1 · 2017-11-15T02:09:32

a)

Seperate the variables
$-1/(kx^2)dx = dt$

Take integral of both sides
$\int-1/(kx^2)dx = \intdt$

$1/(kx)+C = t$

Solve for x
$1/(kx) = t-C$

$kx = 1/(t-C)$

#x = 1/(k(t-C))

b)

Plugin both initial conditions to get two equations
$1 = 1/(k(0-C))$
$0.5 = 1/(k(2-C))$

Simplify both equations
$k = -1/C$
$k(2-C) = 2$

Solve for k and C using substitution
$-2/C+1=2$
$-2/C=1$
$C = -2$
$k = -1/-2=1/2$

c)

Plug values of k and C back into equation for x

$x = 1/(1/2(t-(-2)))$

Simplify

$x = 2/(t+2)$

Find time where x reaches 0.1

$0.1 = 2/(t+2)$
$0.1(t+2)=2$
$t+2=20$
$t=18$