Answer 1

\(15ph=\dfrac{1}{4}h\)

a) \(\left\{{}\begin{matrix}t_1=\dfrac{S_1}{v}=\dfrac{6}{v}\\t_2=\dfrac{S_2}{v}=\dfrac{6+\dfrac{1}{4}.6}{v}=\dfrac{\dfrac{49}{4}}{v}\end{matrix}\right.\)

\(\Rightarrow t_2-t_1=\dfrac{\dfrac{49}{4}}{v}-\dfrac{6}{v}=\dfrac{1}{4}\)\(\Rightarrow\dfrac{\dfrac{25}{4}}{v}=\dfrac{1}{4}\Rightarrow v=25\left(\dfrac{km}{h}\right)\)

b) \(t_1=t_2\Leftrightarrow\dfrac{S_1}{v_1}=\dfrac{S_2}{v_2}\Leftrightarrow\dfrac{6}{25}=\dfrac{6+\dfrac{1}{4}.6}{v_2}\)

\(\Leftrightarrow v_2=\dfrac{\left(6+\dfrac{3}{2}\right).25}{6}=31,25\left(\dfrac{km}{h}\right)\)