Problema 12 si 13 va rog.
problema 12 incepe asa : un mobil a parcurs jumatate din distanta d=450 in t1= 25 s si cealalta jumatate in t2=20s. aflati
acceleratia mobilului
viteza initiala
viteza medie

12.

d₁=d/2

d₂=d/2

d₁+d₂=d

t=t₁+t₂

d₁=d/2=v₀t₁+at₁²/2 (*)

d₂=d/2=v₁t₂+at₂²/2 (1)

v₁=v₀+at₁ (2)

Din (1) si (2)=>d/2=v₀+at₁)t₂+at₂²/2 (**)

Din (*) si (**)=>a=d(t₁-t₂)/[t₁t₂(t₁+t₂)]=450(25-20)/[25*20(25+20)]=450*5/(500*45)=0,1 m/s²

v₀=(d-at₁²)/2t₁=(450-62,5)/50=7,75 m/s

vm=d/t=450/(25+20)=10 m/s

13.

vm=108 km/h=30 m/s

vm=at₁=>t₁=vm/a=30/1=30 s

d₁=at₁²/2=1*900/2=450 m

t'=t/2=2/2=1 min=60 s

d₂=vm*t'=30*60=1800 m

d=2d₁+d₂=900+1800=2700 m