Part A
Given that [tex]a= \frac{dv}{dt} =-kv^2[/tex]
Then,
[tex] \int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c[/tex]
For [tex]v(0)=v_0[/tex], then
[tex]v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0[/tex]
Thus, [tex]v(t)=-kv(t)^2t+v_0[/tex]
For [tex]v(t)= \frac{1}{2} v_0[/tex], we have
[tex] \frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0} [/tex]
Part B
Recall that from part A,
[tex]v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a[/tex]
Now, at initial position, t = 0 and [tex]v=v_0[/tex], thus we have
[tex]x=a[/tex]
and when the velocity drops to half its value, [tex]v= \frac{1}{2} v_0[/tex] and [tex]t= \frac{2}{kv_0} [/tex]
Thus,
[tex]x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a[/tex]
Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by
[tex]- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k} [/tex]
