Question

While vacationing in the mountains you do some hiking. In the morning, your displacement is S⃗ morning= (2200 m , east) + (4000 m north) + (100 m , vertical). Continuing on your hike after lunch, your displacement is S⃗ afternoon= (1300 m , west) + (2500 m , north) - (300 m , vertical).At the end of the hike, how much higher or lower are you compared to your starting point? What is the magnitude of your net displacement for the day?

Answer 1

a

The hiker (you ) is 200 m below his/her(your) starting point

b

The resultant displacement in the north east direction is

`a  = 6562.0 \  m`

The resultant displacement in vertical direction (i.e the altitude change )

`b =6503.1 \  m`

Step-by-step explanation:

From the question we are told that

The displacement in the morning is
`S_(morning) =  (2200 \m , east) + (4000\ m\ north) + (100 \ m ,\ vertical)`

The displacement in the afternoon is
`S _(afternoon)= (1300\ m ,\ west) + (2500 \ m ,\ north) - (300\ m ,\ vertical)`

Generally the direction west is negative , the direction east is positive

the direction south is negative , the direction north is positive

resultant displacement is mathematically evaluated as

`(2200 \m , east) +( - 1300\ m ,\ west) = 900 \ m \ east`

`(4000\ m\ north)  + (2500 \ m ,\ north) = 6500  \ m ,\ north`

`(100 \ m ,\ vertical) - (300\ m ,\ vertical) = -200 \ m`

From the above calculation we see that at the end of the hiking the hiker (you) is 200 m below his/her(your) initial position

Generally from Pythagoras theorem , the resultant displacement in the north east direction is

`a  =  √(900^2 + 6500^2)`

=>
`a  = 6562.0 \  m`

Generally from Pythagoras theorem , the resultant displacement in vertical direction (i.e the altitude change )

`b = √(6500^2 +(-200)^2  )`

=>
`b =6503.1 \  m`