Question

Question #0ca28

Answer 1

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`