$a=-0.05v^2\[2ex]\therefore v\dfrac{dv}{dx}=-0.05v^2\implies \dfrac{dv}{v}=-0.05dx\[2ex]\therefore \log v=-0.05x+c.....(1)$   at x=0;v=20.....put in (1) $\therefore C=\ln(20)=3$   (i) x=?;v=15 $\log v=-0.05\times x+3\[2ex]\therefore \dfrac{\log 15-3}{-0.05}=x\implies x=5.84m$   (ii) a=?;x=50 $\log v =-0.05\times x+3\[2ex]\therefore v=e^{-0.05\times 50+3}=e^{0.5}=1.65m/s\[2ex]a=-0.05v^2=-0.05\times 1.65^2=-0.14m/s^2$