Question

Vector A with arrow, which is directed along an x axis, is to be added to vector B with arrow, which has a magnitude of 9.0 m. The sum is a third vector that is directed along the y axis, with a magnitude that is 3.0 times that of A with arrow. What is the magnitude of A with arrow? (Enter your answer to 4 significant figures.)

Answer 1

2.846m

Step-by-step explanation:

We need to separate each component, and we will write them in the form (x,y), so each vector is:

A: (Ax, Ay)

B: (Bx,By)

A+B: (Ax+Bx, Ay+By).

We know that A is along the x axis, so Ay=0.

We know that A+B is in the y axis, so Ax+Bx=0, which means Bx=-Ax.

For now then we have:

A: (Ax,0)

B: (-Ax,By)

A+B: (0,By)

Clearly the magnitude of A is Ax, and the magnitude of A+B is By (we could use Pythagoras but since it is not necessary we will reserve that for B), and since we know that A+B magnitude is 3 times that of A we know that By=3Ax. This means then that:

B: (-Ax,3Ax).

We now calculate the magnitude of B using Pythagoras:

`|B|=√(B_x^2+B_y^2)=√((-A_x)^2+(3A_x)^2)=√(A_x^2+9A_x^2)=√(10A_x^2)=√(10)A_x`

And we know that this must be 9m

`√(10)A_x=9m`

Which means

`A_x=(9m)/(√(10))=2.84604989415m`

But we also know that the magnitude of A is Ax, so (with 4 significant figures):

`|A|=(9m)/(√(10))=2.846m`