Răspuns:
Rezolvare 1
F=ma ⇒ a = F/m
v= v₀+a·Δt
in intervalul 0s < t < 4s corpul are o miscare uniform accelerata, a = 4/4 = 1 m/s²
viteza la momentul t = 4 s este: v₁ = 0 + 4/4·(4-0) = 4 m/s
in intervalul 4s < t < 6s corpul are o miscare accelerata neuniform, forta si acceleratia medie fiind
[tex]F_{m} =(4-0)/2 = 2 N\\a_{m} = F_{m}/4 = 1/2 m/s^{2}[/tex]
viteza la momentul t = 6 s este: v₂= v₁+[tex]a_{m}[/tex]·Δt = 4 + (1/2)·(6-2) = 5 m/s
Rezolvare 2
[tex]F(t)= \left \{ {{4 N, 0 s \leq t < 4 s} \atop {-2x+12, 4 s \leq t \leq 6 s}} \right.[/tex]
[tex]F = ma => a(t) = F(t)/m =\left \{ {{1 , 0 s \leq t < 4 s} \atop {-\frac{1}{2} x+3, 4 s \leq t \leq 6 s}} \right.[/tex]
[tex]v(t) = v_{0} +\int\limits^t_0 {a(t)} \, dt[/tex]
[tex]v |_{t=6} = v_{0} +\int\limits^6_0 {a(t)} \, dt= 0 +\int\limits^4_0 {1} \, dt+\int\limits^6_4 {(-\frac{1}{2}x+3) } \, dt=[/tex]
= t /₀⁴ - (1/4)·t²/₄⁶+ 3·t /₄⁶ = [tex]4 - 0 - \frac{1}{4}(36-16)+ 3(6-4) = 5 m/s[/tex]
viteza finală este v = 5 m/s
