Question

Intr-o miscare uniform accelerata , in 6 s se parcurge distanta de 15 m , viteza crescand de 4 ori . Cat este acceleratia miscarii ?

Answer 1

[tex]\displaystyle Se~da:\\ \\ t=6s\\ \\ d=15m\\ \\ v=4\times v_0\\ \\ a=?\frac{m}{s^2}\\ \\ \\ Formule:\\ \\ v=v_0+a\times t\\ \\ 4\times v_0=v_0+a\times t\\ \\ 3\times v_0=a\times t\\ \\ v_0=\frac{a\times t}3~(1)\\ \\ \\[/tex]


[tex]\displaystyle d=\frac{\Delta v^2}{2\times a}\\ \\ \Delta v^2=2\times d\times a\\ \\ v^2-v_0^2=2\times d\times a\\ \\ (4\times v_0)^2-v_0^2=2\times d\times a\\ \\ 16\times v_0^2-v_0^2=2\times d\times a\\ \\ 15\times v_0^2=2\times d\times a\\ \\ v_0=\sqrt\frac{2\times d\times a}{15}~(2)\\ \\ \\[/tex]


[tex]\displaystyle Egalam~(1)~si~(2):\\ \\ \frac{a\times t}3=\sqrt\frac{2\times d\times a}{15}\\ \\ \frac{a^2\times t^2}9=\frac{2\times d\times a}{15}\\ \\ 15\times a^2\times t^2=18\times d\times a~(\div a~\div15)\\ \\ a\times t^2=1,2\times d\\ \\ a=\frac{1,2\times d}{t^2}\\ \\ \\ Calcule:\\ \\ a=\frac{1,2\times 15}{6^2}=0,5\frac{m}{s^2}[/tex]