9. determine the vector that must be added to the sum of 𝐴 + 𝐵 in figure 1 to give a resultant…

Explanation:

Step1: Resolve vectors A and B into components

Let the positive x - axis be towards the East and the positive y - axis be towards the North. For vector $\vec{A}$ with $|\vec{A}| = 5.1$ km and direction $38^{\circ}$ East of North: $A_x=|\vec{A}|\sin38^{\circ}=5.1\times\sin38^{\circ}\approx 3.14$ km (East - direction) $A_y = |\vec{A}|\cos38^{\circ}=5.1\times\cos38^{\circ}\approx 4.02$ km (North - direction) For vector $\vec{B}$ with $|\vec{B}| = 6.8$ km and direction $19^{\circ}$ South of West: $B_x=-|\vec{B}|\cos19^{\circ}=- 6.8\times\cos19^{\circ}\approx - 6.42$ km (West - direction) $B_y=-|\vec{B}|\sin19^{\circ}=-6.8\times\sin19^{\circ}\approx - 2.22$ km (South - direction) The sum of the x - components of $\vec{A}$ and $\vec{B}$ is $R_x=A_x + B_x=3.14-6.42=- 3.28$ km The sum of the y - components of $\vec{A}$ and $\vec{B}$ is $R_y=A_y + B_y=4.02-2.22 = 1.8$ km So, $\vec{R}=\vec{A}+\vec{B}=(-3.28\hat{i}+1.8\hat{j})$ km

Step2: Find the vector for resultant displacement of 0

Let the vector to be added be $\vec{C}=(C_x\hat{i}+C_y\hat{j})$. If $\vec{R}+\vec{C}=\vec{0}$, then $\vec{C}=-\vec{R}$. $C_x = 3.28$ km and $C_y=-1.8$ km $|\vec{C}|=\sqrt{C_x^{2}+C_y^{2}}=\sqrt{(3.28)^{2}+(-1.8)^{2}}\approx\sqrt{10.76 + 3.24}=\sqrt{14}\approx3.74$ km The direction $\theta=\tan^{-1}\left(\frac{|C_y|}{|C_x|}\right)=\tan^{-1}\left(\frac{1.8}{3.28}\right)\approx28.9^{\circ}$ South of East

Step3: Find the vector for resultant displacement of 4.0 km [W]

Let the vector to be added be $\vec{D}=(D_x\hat{i}+D_y\hat{j})$. The desired resultant $\vec{R_d}=(-4.0\hat{i}+0\hat{j})$ km We know that $\vec{R}+\vec{D}=\vec{R_d}$ So, $D_x=R_d_x - R_x=-4.0-(-3.28)=- 0.72$ km $D_y=R_d_y - R_y=0 - 1.8=-1.8$ km $|\vec{D}|=\sqrt{D_x^{2}+D_y^{2}}=\sqrt{(-0.72)^{2}+(-1.8)^{2}}=\sqrt{0.5184 + 3.24}=\sqrt{3.7584}\approx1.94$ km The direction $\theta_D=\tan^{-1}\left(\frac{|D_y|}{|D_x|}\right)=\tan^{-1}\left(\frac{1.8}{0.72}\right)\approx68.2^{\circ}$ South of West

Answer:

(a) The magnitude of the vector is approximately $3.74$ km and direction is approximately $28.9^{\circ}$ South of East. (b) The magnitude of the vector is approximately $1.94$ km and direction is approximately $68.2^{\circ}$ South of West.