Ashish walked 50m towards East and took a right turn and walked 40m. He again…

Concept

In a direction-and-distance problem, resolve the whole path into two perpendicular axes — East-West and North-South. Two legs walked along the same axis in opposite directions with equal length cancel each other completely; whatever remains uncancelled on either axis is the net displacement, and the straight-line distance from the start is the hypotenuse of the two remaining axis values (Pythagoras) when both are non-zero.

Applying it here

1. Take East as the positive x-direction and North as the positive y-direction, with Ashish starting at (0, 0) while facing East.

2. First leg: he walks 50 m East, moving to (50, 0).

3. He turns right — facing East, a right turn points him South — and walks 40 m South, moving to (50, -40).

4. He turns right again — facing South, a right turn points him West — and walks 50 m West, moving to (50 - 50, -40) = (0, -40).

5. Final position (0, -40): the East-West displacement is 0 m (the two 50 m horizontal legs cancel exactly), and the North-South displacement is 40 m.

Cross-check

Since the East-West component is 0, the straight-line distance is just the magnitude of the remaining North-South component: √(02 + 402) = 40 m. This also follows from vector cancellation directly — two equal, opposite horizontal legs must cancel by definition, leaving only the untouched vertical leg.

So, Ashish is 40 m away from his starting point.