Question

A man drives his car 5 km in east wart direction he turned right and went for 3 km then he turned west and drove for 1 km. How far is he from the starting point?

Answer 1

It is easily solved with daigram or vector method.

we know, position x axis ⇒east
negative x - axis ⇒ west
positive y - axis ⇒ North
Negative y - axis ⇒south

Now, man drives car 5 km in East
so, r₁ = 5i
He turned right for 3 km , it means positive y - axis
so, r₂ = 3j
again he turned west for 1 km , it means he turned negative x axis.
so, r₃ = -i

Now, position vector of man = r₁ + r₂ + r₃
= 5i + 3j - i = 4i + 3j
Magnitude of position vector of man = |4i + 3j | = 5 km
Hence, man is 5 km far from the starting point.

Answer 2

As per as the question, Displacement covered by the man in east direction is 5 km. 

Hence, In vector notation, Displacement covered by the man in x-axis ($\vec{d_1}$) = 5 $\hat{i}$

Now, Displacement covered in right (north or y-axis) = 3 km.
∴ In vector notation displacement covered ( $\vec{d_2}$ )= 3 $\hat{j}$

Displacement covered in west, or negative x axis = 1 km.
∴ In vector notation, Displacement ( $\vec{d_3}$ )  = - $\hat{i}$


Thus, Displacement Vector = $\vec{d_1} + \vec{d_2} + \vec{d_3}$
  =  $5 \hat{i} + 3 \hat{j} + ( - \hat{i})$
  = $4 \hat{i} + 3 \hat{j}$
  = $\sqrt{(4)^2 + (3)^2}$
 = $\sqrt{25}$
 = 5 km. 


Hence, the displacement (or distance) at which the man is from the initial position is 5 km. 



Hope it helps.