Question

A helicopter goes straight up 500m from a landing pad. It then goes north 20m. Then it goes down 452m. a) What is the displacement of the helicopter?
Express as components of a vector.
x-component_____________________
y-component_____________________

b) What is the displacement of the helicopter? Express as a vector (magnitude and direction).

Answer_____________________

Answer 1

Answer:

a

  x-component             $20 \ m$

y-component     $500 - 452 = 48 \ m$

b

 Magnitude $d = 52 \ m$

direction is  $\theta = 67.4^o$

Explanation:

From the question we are told that

   The first  vertical distance is  $y_1 = 500 \ m$

    The  first horizontal distance  is  $x = 20 \ m$

    The  second vertical distance is  $y_2 = 452 \ m$

Generally the displacement is  

x-component             $20 \ m$

y-component     $500 - 452 = 48 \ m$

Generally the helicopters displacement is mathematically evaluated as  

       $d = \sqrt{ x- component ^2 + y- component ^2 }$

      $d = \sqrt{ 20t ^2 + 48 ^2 }$

      $d = 52 \ m$

The  direction is the angle the displacement of the helicopter makes with the horizontal which is mathematically evaluated as

         $\theta = tan ^{-1}[ \frac{48}{20}]$

=>       $\theta = tan ^{-1}[ 2.4 ]$

=>      $\theta = 67.4^o$