Question

can anyone help me with this2 physics questions

Answer 1

$a = b {v}^{2} \\ \frac{dv}{dt} = - b {v }^{2} \\ \binom{v}{2} ln( \frac{dv}{{v}^{2} } ) = \binom{2}{0} ln( - bdt) \\ - \frac{1}{v} - \frac{1}{2} = - 2b \\ \frac{1}{v} = \frac{4b + 1}{2} \\ \frac{dx}{dt} = \frac{2}{4b + 1} \\ ln(dx) = ln( \frac{2}{4b + 1} ) dt \\ x = ln( \frac{4b + 1}{b} )$