Question

A student walks 50 meters east, 40 meters north, 35 meters east, and then 20 m south. What is the magnitude and direction of the student's total displacement

Answer 1

Answer: A student walks 50 meters east, 40 meters north, 35 meters east, and then 20 m south. Then the magnitude and direction of the student's total displacement will be 87.32 m along the direction of AD or in east-south direction.

Explanation: To find the correct answer, we need to know about the Displacement of a body in motion.

What is displacement of a body in motion?

• The displacement is the shortest distance between initial and final positions of a body.
• It's a vector quantity, and can positive, negative, or zero.
• The magnitude of displacement is less than or equal to the distance travelled.

How to solve the problem?

• At first, we can draw a diagram showing the motion of the body.
• From the diagram, the displacement of the body will be equal to the distance between point A and D.
• To solve this, we can use Pythagoras theorem.

$AD=AC+CD\\AC^{2} =50^{2} +11^2\\AC=51.19 m\\Similarly,\\CD^2=35^2+9^2\\CD=36.13 m\\thus, \\AD=51.19+36.13=87.32 m$

Thus, from the above calculations, we can conclude that, the displacement of the body will be equal to 87.32 m along the direction of AD or in east-south direction.

Learn more about the Displacement here:

brainly.com/question/28020108

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