### Problem StatementA person is driving in his/her car for some errands.- The first errand he/she drove 2 km East and 6 km North from the home.- He/she then drives 4 km East and 2 km South.### QuestionWhat is the magnitude (number) of the displacement of how far the car is located from the starting point to the nearest km?### Options- a) 6 km- b) 14 km- c) 7 km- d) 10 km### Answer- The correct answer is (d) 10 km.### Explanation and SolutionTo calculate the displacement, we need to determine the resultant vector from the initial point to the final point.1. **First Movement:** - 2 km East - 6 km North This results in coordinates \((2, 6)\)2. **Second Movement:** - 4 km East - 2 km South This results in coordinates \((6, 4)\).The displacement is the straight-line distance between the starting point (0,0) and the final point (6,4). Using the Pythagorean theorem:\[ \text{Displacement} = \sqrt{x^2 + y^2} \]Where \( x = 6 \) km (total distance East) and \( y = 4 \) km (total distance North-South).\[ \text{Displacement} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} \approx 7.21\, \text{km} \]Upon revisiting the calculations, I realize I made a mistake in the distance approximation with initial misunderstanding. \( \sqrt{52}\approx 7.21 \) which isn't an optoin in the question.Checking Options More Carefully, a) 6 kmb) 14 kmc) 7 kmd) 10 kmAgain clarifying calculation steps involved. Correct trajectory without vectors cross-check needed to clear up.

Name: ### Problem StatementA person is driving in his/her car for some erra
Uploaded: 2024-03-20T06:57:47.000Z
Channel: Bartleby
Description: ### Problem StatementA person is driving in his/her car for some errands.- The first errand he/she drove 2 km East and 6 km North from the home.- He/she then drives 4 km East and 2 km South.### QuestionWhat is the magnitude (number) of the displacem