Explanation:
Given: a = -3v^2
By definition, the acceleration is the time derivative of velocity v:
[tex]a = \frac{dv}{dt} = - 3 {v}^{2} [/tex]
Re-arranging the expression above, we get
[tex] \frac{dv}{ {v}^{2} } = - 3dt[/tex]
Integrating this expression, we get
[tex] \int \frac{dv}{ {v}^{2} } = \int {v}^{ - 2}dv = - 3\int dt[/tex]
[tex] - \frac{1}{v} = - 3t + k[/tex]
Since v = 10 when t = 0, that gives us k = -1/10. The expression for v can then be written as
[tex] - \frac{1}{v} = - 3t - \frac{1}{10} = - ( \frac{30 + 1}{10} )[/tex]
or
[tex]v = \frac{10}{30t +1 } [/tex]
We also know that
[tex]v = \frac{ds}{dt} [/tex]
or
[tex]ds = vdt = \frac{10 \: dt}{30t + 1} [/tex]
We can integrate this to get s:
[tex]s = \int v \: dt = \int ( \frac{10}{30t + 1}) \: dt = 10 \int \frac{dt}{30t + 1} [/tex]
Let u = 30t +1
du = 30dt
so
[tex] \int \frac{dt}{30t + 1} = \frac{1}{30} \int \frac{du}{u} = \frac{1}{30}\ln |u| + k[/tex]
[tex]= \frac{1}{30}\ln |30t + 1| + k[/tex]
So we can now write s as
[tex]s = \frac{1}{3}\ln |30t + 1| + k[/tex]
We know that when t = 0, s = 8 m, therefore k = 8 m.
[tex]s = \frac{1}{3}\ln |30t + 1| + 8[/tex]
Next, we need to find the position and velocity at t = 3 s. At t = 3 s,
[tex]v = \frac{10}{30(3) +1 } = \frac{10}{91}\frac{m}{s} = 0.11 \: \frac{m}{s} [/tex]
[tex]s = \frac{1}{3}\ln |30(3) + 1| + 8 = 9.5 \: m[/tex]
Note: velocity approaches zero as t --> [tex]\infty [/tex]
