Given that

$a=-0.05v^2 --------- (i)$

We know that,

$v=\dfrac {dx}{dt} \\ a=\dfrac {dv}{dt} \\ \therefore \dfrac va= \dfrac {dx}{dv} \\ \therefore \dfrac {vdv}{dx}=a \\ \therefore \dfrac { vdv}{dx}=-0.05v^2 \\ \therefore dx=-\dfrac {dv}5 = -\dfrac {20 dv}v ------------(ii)$

At v=20m/s, x=0. Hence x at 15m/s is given by

$x=-∫\limits_{20}^{15}\dfrac {dv}{0.05v } =-20 ln⁡(\dfrac {15}{20}) \\ \therefore x=5.754m\space\space \space at\space\space \space v=15m/s \Rightarrow Ans$

Also from (ii):-

$∫\limits_0^{50} dx=∫\limits_{20}^v -\dfrac {20dv}v \\ \therefore 50=-20 ln⁡(\dfrac v{20}) \\ \therefore \dfrac v{20}=e^{-5/2} \\ \therefore v=20e^{-5/2} (at\space \space x=50m) \\ \therefore a=-0.05(v)^2 \\ =-0.05(20e^{-5/2})^2 \\ \therefore a=-0.135m/s^2 (at \space \space x=50m) -- Ans$