Solving the Distance and Direction Problem

This problem asks us to find the final position of the bus stand relative to the railway station, given a sequence of Mohan's movements. We need to track his movement in the East-West and North-South directions separately.

Breaking Down Mohan's Journey

Mohan starts at the railway station. Let's consider the railway station as the origin point (0,0). We will use positive values for East and North, and negative values for West and South.

• Step 1: Walked 120 m towards the east.
• Step 2: Took a right turn and walked another 100 m. From East, a right turn is South. So, 100 m South.
• Step 3: Then took a left turn and walked 30 m. From South, a left turn is East. So, 30 m East.
• Step 4: Again, he took a left turn and walked 140 m. From East, a left turn is North. So, 140 m North.
• Step 5: From there, finally took a left turn and walked 150 m to reach the bus stand. From North, a left turn is West. So, 150 m West.

Analyzing Movements by Direction

Let's consolidate the movements in the East-West and North-South directions:

Calculating Net Displacement

Now, let's calculate the total displacement in each primary direction:

• Net East-West movement: +120 ($\text{m}$) (East) + 30 ($\text{m}$) (East) - 150 ($\text{m}$) (West) = 150 ($\text{m}$) - 150 ($\text{m}$) = 0 ($\text{m}$).
• Net North-South movement: -100 ($\text{m}$) (South) + 140 ($\text{m}$) (North) = 40 ($\text{m}$).

The net displacement is 0 ($\text{m}$) in the East-West direction and +40 ($\text{m}$) in the North-South direction. A positive value in the North-South calculation means the final position is North of the starting point.

Determining Final Distance and Direction

The net displacement is 0 ($\text{m}$) horizontally and 40 ($\text{m}$) vertically upwards (North). Therefore, the bus stand is located directly North of the railway station.

The straight-line distance from the railway station to the bus stand is the magnitude of this net displacement.

Distance = $\sqrt{(\text{Net East-West})^2 + (\text{Net North-South})^2}$

Distance = $\sqrt{(0 \, \text{m})^2 + (40 \, \text{m})^2}$

Distance = $\sqrt{0 + 1600 \, \text{m}^2}$

Distance = $\sqrt{1600 \, \text{m}^2}$

Distance = 40 $\text{m}$.

The net displacement is 40 $\text{m}$ in the North direction.

Conclusion

The bus stand is 40 $\text{m}$ away from the railway station in the North direction.

Revision Table: Key Movement Analysis

The final position relative to the start is 0 $\text{m}$ East/West and 40 $\text{m}$ North. This confirms the bus stand is 40 $\text{m}$ North of the railway station.

Additional Information on Direction Problems

Direction and distance problems often involve tracking movements on a 2D plane. The key is to break down each movement into its components along the cardinal directions (North, South, East, West). Turns (left or right) change the current direction of movement.

• A right turn means turning 90 degrees clockwise from the current direction.
• A left turn means turning 90 degrees anti-clockwise from the current direction.

By summing the total displacements in the East-West axis and the North-South axis separately, you can find the net change in position. The final distance is the hypotenuse of a right triangle formed by the net East-West and net North-South displacements (using Pythagoras theorem). The direction is determined by the quadrant of the final net displacement vector relative to the starting point.

In this specific problem, the net East-West displacement is zero, simplifying the calculation of the final distance and direction significantly.