Question Number 181954 by yaga7 last updated on 02/Dec/22
Answered by Acem last updated on 02/Dec/22
Answered by Ritzy last updated on 02/Dec/22
$$\mathrm{let}\:\mathrm{the}\:\mathrm{first},\:\mathrm{the}\:\mathrm{secon}\bar {\mathrm{d}}\:\mathrm{and}\:\mathrm{the}\:\mathrm{third}\:\mathrm{displacement}\:\mathrm{be}\:\overset{\rightarrow} {\mathrm{A}},\:\overset{\rightarrow} {\mathrm{B}}\:{and}\:\overset{\rightarrow} {\mathrm{C}}\:\mathrm{respectively} \\ $$
Commented by Acem last updated on 02/Dec/22
$${ofcourse} \\ $$
Answered by Ritzy last updated on 02/Dec/22
$$\mathrm{let}\:\mathrm{the}\:\mathrm{first},\:\mathrm{the}\:\mathrm{second}\:\mathrm{and}\:\mathrm{the}\:\mathrm{third}\:\mathrm{position}\:\mathrm{be}\:\overset{\rightarrow} {\mathrm{A}},\:\overset{\rightarrow} {\mathrm{B}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{C}}\:\mathrm{respectively} \\ $$$$\overset{\rightarrow} {\mathrm{A}}\:=\:\mathrm{2}\overset{\rightarrow} {\mathrm{i}},\:\overset{\rightarrow} {\mathrm{B}}\:=\:\mathrm{3cos40}^{\mathrm{0}} \overset{\rightarrow} {\mathrm{i}}\:+\:\mathrm{3sin40}^{\mathrm{0}} \overset{\rightarrow} {\mathrm{j}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{C}}\:=\:\mathrm{2}.\mathrm{5}\overset{\rightarrow} {\mathrm{j}} \\ $$$$\mathrm{R}\:=\:\overset{\rightarrow} {\mathrm{A}}\:+\:\overset{\rightarrow} {\mathrm{B}}\:+\:\overset{\rightarrow} {\mathrm{C}} \\ $$