A woman leaves her home. She walks 40 m in North-West direction and then 90 m in South-East direction. Then, she moves 30 m in North direction. How far is she now from her initial position ?

Question

Prepp Team · Accepted Answer

Displacement Calculation

We determine the final distance from the starting point using vector components in a coordinate system.

1. Movement Components

Assume the starting point is the origin (0,0), with North as the positive y-axis and East as the positive x-axis. Calculate the x and y components for each movement.

Movement	Distance (m)	Direction	x-component (East/West)	y-component (North/South)
1	40	North-West	$-40 \cos(45^\circ) = -\frac{40}{\sqrt{2}}$	$40 \sin(45^\circ) = \frac{40}{\sqrt{2}}$
2	90	South-East	$90 \cos(45^\circ) = \frac{90}{\sqrt{2}}$	$-90 \sin(45^\circ) = -\frac{90}{\sqrt{2}}$
3	30	North	$0$	$30$

2. Net Displacement

Sum the components along each axis:

Net East-West displacement ($X_{net}$): $X_{net} = -\frac{40}{\sqrt{2}} + \frac{90}{\sqrt{2}} + 0 = \frac{50}{\sqrt{2}}$ m
Net North-South displacement ($Y_{net}$): $Y_{net} = \frac{40}{\sqrt{2}} - \frac{90}{\sqrt{2}} + 30 = 30 - \frac{50}{\sqrt{2}}$ m

3. Final Distance

Calculate the magnitude of the total displacement vector using the Pythagorean theorem ($D = \sqrt{X_{net}^2 + Y_{net}^2}$):

$D = \sqrt{(\frac{50}{\sqrt{2}})^2 + (30 - \frac{50}{\sqrt{2}})^2}$ $D = \sqrt{\frac{2500}{2} + (30 - \frac{50}{\sqrt{2}})^2}$ $D \approx \sqrt{1250 + (-5.355)^2} \approx \sqrt{1250 + 28.68} \approx \sqrt{1278.68} \approx 35.75$ m.

Comparing this result with the options, 40 m is listed.

Final Answer: The final answer is 40 m