Question

a person moves 3km towards east, 4km towards south, 3.5 km towards east. find the distance covered and the displacement of the person?​

Answer 1

Topic :- Motion in straight line.

$\maltese\:\underline{\sf AnsWer :}\:\maltese$

Let the starting point be "A" . From "A" he moves 3 km upto point "B" towards east . And from point "B" he moves 4 km towards south at point "D". and again he moves 3.5 from point "D" towards "C" in direction east. For more clear visualisation I've attached the figure check it out.

$\dag\:\underline{\tt Total \: Distance \: covered \: by \: the \: person :}$

$\longrightarrow\:\:\sf Total \: distance \: covered = AB + BD + DC$

$\longrightarrow\:\:\sf Total \: distance \: covered = 3 + 4 +3.5$

$\longrightarrow\:\:\sf Total \: distance \: covered = 7 +3.5$

$\longrightarrow\:\: \underline{ \boxed{\sf Total \: distance \: covered =10.5 \: km}}$

$\dag\:\underline{\tt Total \: Displacement \:of the\: the\: person :}$

Displacement means Change in position of the object from initial point to final point. So, from figure we observe that Displacement of the person is from point A to C i.e Displacement = AC

If AB = 3.0 km then point OD = 3.0 km. Also if point BD = 4 km then point AO = 4 km.

By Pythagoras theorem we can find the displacement of the person :

$\dashrightarrow\:\:\sf (Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2$

$\dashrightarrow\:\:\sf (AC)^2 = (OC)^2 + (AO)^2$

$\dashrightarrow\:\:\sf (AC)^2 = (OD + DC)^2 + (AO)^2$

$\dashrightarrow\:\:\sf (AC)^2 = (3 + 3.5)^2 + (4)^2$

$\dashrightarrow\:\:\sf (AC)^2 = (6.5)^2 + (4)^2$

$\dashrightarrow\:\:\sf (AC)^2 = 42.25 + 16$

$\dashrightarrow\:\:\sf (AC)^2 = 582.5$

$\dashrightarrow\:\:\sf AC = \sqrt{582.5 }$

$\dashrightarrow\:\: \underline{ \boxed{\sf AC = 76.3 \: km}}$