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

Answer:

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 )

  $b =6503.1 \ m$

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

      $a = \sqrt{900^2 + 6500^2}$

=>     $a = 6562.0 \ m$

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

      $b = \sqrt{6500^2 +(-200)^2 }$

=>   $b =6503.1 \ m$