Question

Un corp, pornind din repaus, se mişcă rectiliniu uniform accelerat. În secunda k = 3 parcurge
distanţa d = 5 cm. Viteza v la momentul τ = 12 s este?

Answer 1

[tex]\displaystyle Se~da:\\ \\ k=3s\\ \\ d=5cm=0,05m\\ \\ \tau=12s\\ \\ v=?\frac ms\\ \\ \\ Formule:\\ \\ In~a~k~secunda~corpul~parcurge:\\ \\ d=v_0\times 1+\frac{a\times 1^2}{2}=v_0+\frac a2\\ \\ \\ v_0~este~viteza~la~sfirsitul~secundei~2(k-1)\\ \\ v_0=a\times (k-1)\\ \\ \\[/tex]


[tex]\displaystyle Inlocuind:\\ \\ d=a\times(k-1)+\frac a2\\ \\ d=a\times(k-1+0,5)=a\times(k-0,5)\\ \\ a=\frac{d}{k-0,5}\\ \\ \\ v=a\times \tau=\frac{d\times \tau}{k-0,5}\\ \\ \\ Calcule:\\ \\ v=\frac{0,05\times 12}{3-0,5}=0,24\frac ms[/tex]