\(t_1=\dfrac{1s}{4}.\dfrac{1}{v_1}\left(h\right)\\ t_2=\dfrac{3s}{4}.\dfrac{1}{v_2}\left(h\right)\)
Ta có :
Tốc độ trung bình là:
\(v_{tb}=\dfrac{s}{t_1+t_2}=\dfrac{s}{\dfrac{s}{4}.\dfrac{1}{v_1}+\dfrac{3s}{4}.\dfrac{1}{v_2}}=\dfrac{1}{\dfrac{1}{4v_1}+\dfrac{3}{4v_2}}=\dfrac{1}{\dfrac{1}{4.30}+\dfrac{3}{4.60}}=48\)(km/h)
Đáp án B