Question

A child walks 5.0 meters north, then 4.0 meters east, and finally 2.0 meters south. what is the magnitude of the resultant displacement of the child after the entire walk

Answer 1

The magnitude of the resultant displacement of the child after walking 5.0 meters north, 4.0 meters east, and 2.0 meters south is 5.0 meters, calculated using the Pythagorean theorem.

Step-by-step explanation:

To find the magnitude of the resultant displacement of the child who walks 5.0 meters north, then 4.0 meters east, and finally 2.0 meters south, we can use the concept of vector addition. First, we must break down the walk into its north-south and east-west components. The child walks a net distance of 5.0 meters north minus 2.0 meters south, which equals 3.0 meters north. There is no subtraction for the eastward walk as there is no westward component, so it remains 4.0 meters east.

Next, we use the Pythagorean theorem to calculate the resultant displacement. This is the hypotenuse of a right triangle formed by the north-south and east-west displacements as the two legs:

R = √(3.0 m)^2 + (4.0 m)^2
√9 + 16
√25
R = 5.0 meters

Therefore, the magnitude of the resultant displacement is 5.0 meters.

Answer 2

Resultant displacement of the child is 5 m.

Step-by-step explanation:

It is given that, a child walks 5.0 meters north, then 4.0 meters east, and finally 2.0 meters south.

We have to find the magnitude of displacement of the child after the entire walk. The attached figure shows entire journey of the child. Displacement of an object is equal to the shortest path covered by it. Here, AM shows the net displacement. It can be calculated using Pythagoras theorem as :

Since, AB = 5 m So, OA = 3 m and BC = OM = 4 m

`OA^2+OM^2=AM^2`

`3^2+4^2=AM^2`

⇒ AM = 5 m

Hence, the resultant displacement of the child after the entire walk is 5 m.