Câu 3

Theo đồ thi \(B_1\&B_2\) là chuyển động thẳng nhanh dần đều

\(\left\{{}\begin{matrix}v_1=v_{o1}+a_1t\\v_2=v_{o2}+a_2t\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}s_1=v_{o1}t+\dfrac{1}{2}a_1t^2\\s_2=v_{o2}t+\dfrac{1}{2}a_2t^2\end{matrix}\right.\)

mà \(v_{o1}=v_{o2}=0;a_1=tan60^o=\sqrt{3};a_2=tan30^o=\dfrac{1}{\sqrt{3}}\)

\(\Rightarrow\dfrac{s_2}{s_1}=\dfrac{a_2}{a_1}=\dfrac{\dfrac{1}{\sqrt{3}}}{\sqrt{3}}=\dfrac{1}{3}\approx0,3\)