Question

Mohit was looking for his father . He went 90 metres in the east before turning to his right . He went 20 metres before turning to his right again to look for his father his uncle's place 30 metres from this point . His father was not there . From here he went 100 metres to the north before meeting his father in a street . how far did the son meet his father from the starting point ​

Answer 1

Let i, j be unit vectors in the direction of East and North respectively, S represent the distance.

Mohit went 90 meters in the east

⇒ S= 90i + 0j

He turns right and travels 20 meters

⇒ S = - 20 j

He again turns right and travels 30 meters to his uncle's place

⇒ S = - 30 i

From here, He travels 100 m to the north where he meets his father.

⇒ S = 100 j

Now, Total Displacement is given :

⇒S = 90 i - 20 j - 30 i + 100 j

⇒ S = 60i + 80 j

Distance traveled by the boy from the starting point to the place where he met his father is :

⇒ |S| = 100 m.

Therefore, The son met his father 100 m away from the starting point.

Answer 2

Clearly, the child moves from A to B 90 metres eastwards upto B, then turns right and moves 20 metre upto C, then turns right and moves upto 30 metre upto D. Finally he turns right and moves upto 100 metre upto E.

So AB = 90 metre, BF = CD = 30 metre,

So, AF = AB - BF = 60 metre

Also DE = 100 metre, DF = BC = 20 metre

So, EF = DE - DF = 80 metre

as we can see in image that triangle AFE is a right angled triangle and we are having two sides, need to calculate third one, so we can apply Pythagoras theorem here

\begin{aligned}

A = AE = \sqrt{AF^2 + EF^2} \\

= \sqrt{(60)^2+(80)^2} \\

= \sqrt{3600+6400} \\

= \sqrt{10000} = 100

\end{aligned}

So from starting point his father was 100 metre away.