Question

anita travels 4m towards north, 3m towards east and then 7m towards south. find the displacement and also the distance covered by anita.​

Answer 1

Explanation:

Before commencing the steps, take a look at the second attachment for better understanding.

• AB is distance covered when she travels due North.
• BC is distance covered when she travels due East.
• CD is distance covered when she travels due South.

$\underline{\small \sf {\maltese \; \; \; Finding \: distance \: travelled : \; \; \; }}$

Distance travelled is the total distance covered by the body. It is a scalar quantity and its SI unit is m (metre). So,

$\longrightarrow\tt{Distance = AB + BC + CD } \\ \\ \longrightarrow\tt{Distance = (4 + 3+ 7) \; m } \\ \\ \longrightarrow \underline{\boxed{\tt{Distance = 14 \; m }}} \; \red{\bigstar}$

∴ The distance travelled is 14 m.

$\underline{\small \sf {\maltese \; \; \; Finding \: displacement \: : \; \; \; }}$

Displacement is the shortest distance from initial to final position of the body. Here, initial position is A and the final position is D. So, the displacement is AD. To calculate AD, we need to make so construction in the diagram.

Construction : Make a line AO (imaginary line).

Now, the figure has been split up into two polygons , a right angled triangle and a rectangle. Here,

• BC = AO [3 m] (Opposite angles of a rectangle are equal.) __(1)
• BA = CO [4m] (Opposite angles of a rectangle are equal.)

Since, BA = CO (3 m) So,

• CD = CO + OD
• 7 m = 4 m + OD
• 3m = OD___(2)

Now, in the ∆ AOD :

$\longrightarrow\tt{Displacement = \sqrt{(AO)^2 + (OD)^2 }} \\ \\ \longrightarrow\tt{ Displacement = \sqrt{(3 \; m)^2 + (3 \; m)^2} } \\ \\ \longrightarrow\tt{ Displacement = \sqrt{9 \; m^2 + 9 \; m^2} }\\ \\\longrightarrow\tt{ Displacement = \sqrt{18 \; m^2} } \\ \\ \longrightarrow \underline{\boxed{\tt{Displacement = 3\sqrt{2} \; m }}} \; \red{\bigstar}$

∴ The displacement is 3√2 m towards south east.