Kinematics: Net displacement

[emilyjbee]

A student runs 30m east, 40m north and 50m west. What is the magnitude of the students net displacement? What is the net displacement?

I can't find an equation to solve this problem. Could someone please give me some guildlines to figure out net displacement of multiple directions?

 

[learningphysics]

A student runs 30m east, 40m north and 50m west. What is the magnitude of the students net displacement? What is the net displacement?

I can't find an equation to solve this problem. Could someone please give me some guildlines to figure out net displacement of multiple directions?

Can you simplify the sum of these 3 vectors to a sum of 2... Hint: what is the net displacement of 30m east and 50m west.

 

[m2003]

Sure, I can provide some guidelines for solving this problem. Kinematics is the branch of physics that deals with the motion of objects without considering the cause of the motion. Net displacement is the total distance and direction an object has traveled from its starting point to its end point.

To solve this problem, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. In this case, the hypotenuse represents the net displacement.

First, we can draw a diagram to represent the student's movements. The student runs 30m east, 40m north, and 50m west. This can be represented as a right triangle with sides of 30m, 40m, and 50m.

Next, we can use the Pythagorean theorem to find the length of the hypotenuse, which represents the net displacement. The equation is c^2 = a^2 + b^2, where c is the length of the hypotenuse and a and b are the lengths of the other two sides. Plugging in the values, we get c^2 = 30^2 + 40^2, which simplifies to c^2 = 2500. Taking the square root of both sides, we get c = 50m.

Therefore, the magnitude of the student's net displacement is 50m and the net displacement is 50m in the direction opposite to the direction the student started in (in this case, west).

I hope this helps you understand how to solve this problem using kinematics principles. Remember, always draw a diagram and use the appropriate equations to solve kinematics problems.