A person moves 30m north, then 20m east and finally 30 root 2m south west. Find the displacement of the person.
30√2 SW is the same as 30S and 30W
So, now add up all your distances and see where you end up.
The answer will be surprisingly easy.
zero
@reynis plz mind that lengths are 20w 30n not 30w should it be that it wud have been zero which no one may ask
To find the displacement of the person, we need to consider the net effect of all the movements in terms of both distance and direction.
1. The person moves 30m north.
2. The person then moves 20m east.
3. Finally, the person moves 30√2m (which can be simplified to approximately 42.43m) south-west.
Now, let's break down the movements into their respective components:
- North movement: 30m north (in the positive y-direction)
- East movement: 20m east (in the positive x-direction)
- South-west movement: 30√2m (approximately 42.43m) south-west (which can be divided into a south and west movement)
- South movement: Since the person moved in the south direction, this is considered negative in terms of the y-direction. So, the south movement can be represented by -42.43m in the y-direction.
- West movement: Similarly, the west movement is considered negative in terms of the x-direction. So, the west movement can also be represented by -42.43m in the x-direction.
Now we can compute the net displacement by adding the individual vector components together:
Net displacement in the x-direction = (20m - 42.43m) = -22.43m
Net displacement in the y-direction = (30m - 42.43m) = -12.43m
Using the Pythagorean theorem, we can calculate the magnitude of the net displacement:
Magnitude (displacement) = √((-22.43m)^2 + (-12.43m)^2)
= √(501.49m^2 + 154.36m^2)
= √655.85m^2
= 25.59m (approximately)
Therefore, the displacement (net distance) of the person is approximately 25.59m, and the direction is towards the southwest.