The vector -5.2A? has a magnitude of 47m and points in the positive x direction. Find the x component of the vector A?

Question

I feel like this is an easy question and I am just over thinking it.  A detailed answer, not just the answer would be more helpful.  I also have to find the magnitude of the vector A as well if that could be explained as well.  Thanks!

 

Answer 1

"The vector -5.2A has a magnitude of 47m and points in the positive x direction. Find the x component of the vector A."

 

"The vector -5.2A ... points in the positive x direction":

 

-5.2A = a u + 0 v; where u is the unit vector in the x-direction and v in y-direction (we don't have good symbols).

 

The magnitude of -5.2A, |-5.2A| = √(a^2 + 0^2) = a = 47

 

-5.2A = 47 u + 0 v

 

A = (47/-5.2) u + 0 v

 

"Find the x component of the vector A."

 

47/-5.2 = -9.03846153846154

Answer 2

 If you draw Vector with an angle Θ with horizontal line

 

   X component  of any vector A is A Cos θ :

 

     we have :

 

      X component of - 5.2 A to be 47m

 

      i.e.    - 5.2 A cos θ = 47m

 

 

          A cos Θ ( X component of Vector A ) =

           

             - 47m /5.2  = - 9.04 m