Question

a person moves 30m along north direction, then moves 40m along east direction and finally moves 20m along south direction. Find distance and displacement

Answer 1

Distance = 30 m + 40 m + 20 m.

Displacement Vector = 30 j + 40 i + (-20) j = 40 i + 10 j.
Magnitude of displacement = $\sqrt{40^2+10^2} = 10\sqrt{17}$

Answer 2

Answer:

Distance is equal to 90m and displacement is equal to $10\sqrt{17}$m.

Explanation:

Given , a person moves 30m along north direction  ,

then moves 40m along east direction and finally moves 20m along south direction .

First we need to draw a diagram according to the given information .

So , from the diagram given below ,

AD = 30 m (north direction ),  DC = 40 m  (east direction ) and

CG = 20 m (south direction )

Therefore , BG = CB - CG = 30 - 20 = 10 m

and AC = AB = 40 m .

• Distance is equal to  total distance travel by a body or a person .

⇒Distance = AD + DC +CG

Distance = 30 + 40 + 20 = 90 m

• Now , as we know displacement is just the distance between the object's initial location and its final location.

So from the diagram AG is the required displacement .

Now , in right angle triangle ΔABG ,

AG = $\sqrt{AB^2 +BG^2}$         [By Pythagoras theorem  ]

⇒AG = $\sqrt{40 ^2 + 10^2}$ = $\sqrt{1600 +100}$ = $\sqrt{100(16 + 1)}$

AG = 10$\sqrt{17}$m

Hence , 90m is the value of distance and  displacement is equal to $10\sqrt{17}$ m

#SPJ3