Question

Giúp em gấp ạ

Answer 1

bt1: \(\Rightarrow vtb=\dfrac{2.60+3.40}{5}=48km/h\)

bt2. chon O\(\equiv\)dia diem xe dap xuat phat, chieu(+) la chieu xe dap chuyen dong

-moc tgian luc 7h

\(\Rightarrow\left\{{}\begin{matrix}x1=15t\\x2=10+5t\end{matrix}\right.\) \(\left(km,h,\right)\)gap nhau \(\Rightarrow x1=x2=>t=1h\)

vi tri gap nhau cach A: \(S=x1=15.1=15km\)

bt3: chon \(Ox\equiv AB,O\equiv A,\)moc tgian luc 6h30' 

chieu (+) A->B

a,\(\Rightarrow x1=30t\left(km,h\right)\)

b,\(\Rightarrow t=2,5h\Rightarrow x1=30.2,5=75km\)

=>luc 8h vi tri xe cach A(O) 75km

c, giong y (b) \(\Rightarrow t=\dfrac{75}{30}=2,5h\)