Question
Question: A man walks 30 m towards north, then 20 m towards east and in the last \[30\sqrt 2 \,m\], towards so...
A man walks 30 m towards north, then 20 m towards east and in the last 302m, towards south-west. The displacement from origin is
A. 10 m towards west
B. 10 m towards east
C. 602m towards northwest
D. 602m towards east north
Solution
Use the formula for position vector of a certain point in x-y plane. Calculate the x and y components of south-west displacement of the man. Add the total displacement with respect to its direction.
Formula used:
S=xi^+yj^
Here, x is the x-coordinate and y is the y-coordinate.
Complete step by step answer:
The position vector of point P in x-y plane is given as,
S=xi^+yj^
Here, x is the x-coordinate and y is the y-coordinate.
The displacement of a man is can be drawn as shown in the figure below,
Here, the orange straight-line (d1) represents the northward displacement, yellow straight-line (d2) represents eastward displacement and blue straight-line (d3) represents south-west displacement.
Therefore, the total displacement of a man from origin is,
S=(d2−d3cos45∘)i^+(d1−d3sin45∘)j^
Here, d3cos45∘ is the x-component of south-west displacement and d3sin45∘ is the y-component of south-west displacement,
Substitute 30 m for d1, 20 m for d2 and 302m for d3 in the above equation.
S=(20−302cos45∘)i^+(30−302sin45∘)j^
⇒S=(20−30)i^+0j^
⇒S=−10i^
The negative sign implies that the displacement is along the negative direction that is towards west. Therefore, the net displacement from the origin is 10 m towards the west.
So, the correct answer is “Option A”.
Note:
Always specify the direction of the displacement vector.
In this question, we have taken the east and north as positive directions and west and south as negative directions.