# Vector Displacement| Vector Addition and Subtraction: Graphical Method| Two-Dimensional Kinematics| College Physics Problem 3.7

### (a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0º north of east (which is equivalent to subtracting B from A —that is, to finding R′ =A−B ). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0º south of west and then 12.0 m in a direction 20.0º east of south (which is equivalent to subtracting A from B —that is, to finding R′′ = B – A = – R′ ). Show that this is the case.

SOLUTION:

Part A

We are required to solve for

$\vec{R'}=\vec{A'}+\vec{B}'$

The x-component of the resultant is:

$R_x'=A_x'+B_x'$

$R_x'=12\:cos\left(110^{\circ} \right)+\left(20.0\right)cos\left(40^{\circ} \right)=11.217\:m$

The y-component of the resultant is:

$R_x'=A_y'+B_y'$

$R_x'=\left(12.0\right)sin\left(110^{\circ} \right)+\left(20.0\right)sin\left(40^{\circ} \right)=24.132\:m$

The resultant is

$R'=\sqrt{R_x'+R_y'}$

$R'=\sqrt{\left(11.217\:m\right)^2+\left(24.132\:m\right)^2}$

$R'=26.6\:m$

The compass direction is

$\alpha =tan^{-1}\left(\frac{R_y'}{R_x'}\right)$

$\alpha =tan^{-1}\left(\frac{24.132}{11.217}\right)=65.1^{\circ}$

The compass direction is $65.1^{\circ} \:North\:of\:East$

Part B

The result is the same as that in Part A since the components did not change. Only the order has changed. This is consistent with the fact that $\vec{A}-\vec{B}=-\left(\vec{B}-\vec{A}\right)$