Car Displacement Faa3A8

1. **State the problem:** A car travels $20$ km due north and then $25$ km in a direction $60^\circ$ West of North. We want to find the magnitude and direction of the resultant vector representing the net displacement. 2. **Set up coordinate system and vectors:** - Let north be along the positive $y$-axis. - The first displacement vector is $\vec{A} = 0\hat{i} + 20\hat{j}$ km. - The second displacement vector $\vec{B}$ makes an angle of $60^\circ$ west of north, so its components are: $$B_x = -25 \sin 60^\circ = -25 \times \frac{\sqrt{3}}{2} = -\frac{25\sqrt{3}}{2}$$ $$B_y = 25 \cos 60^\circ = 25 \times \frac{1}{2} = 12.5$$ 3. **Calculate components of $\vec{B}$:** $$B_x = -\frac{25\sqrt{3}}{2} \approx -21.65$$ $$B_y = 12.5$$ 4. **Find resultant vector components:** $$R_x = A_x + B_x = 0 + (-21.65) = -21.65$$ $$R_y = A_y + B_y = 20 + 12.5 = 32.5$$ 5. **Calculate magnitude of resultant vector $\vec{R}$:** $$|\vec{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{(-21.65)^2 + 32.5^2} = \sqrt{469.22 + 1056.25} = \sqrt{1525.47} \approx 39.06$$ 6. **Calculate direction angle $\theta$ of $\vec{R}$ relative to north:** $$\theta = \tan^{-1} \left( \frac{|R_x|}{R_y} \right) = \tan^{-1} \left( \frac{21.65}{32.5} \right) \approx \tan^{-1}(0.666) \approx 33.69^\circ$$ Since $R_x$ is negative and $R_y$ positive, the vector points west of north. 7. **Final answer:** The net displacement vector has magnitude approximately $39.06$ km and direction $33.69^\circ$ West of North. **Graphical method:** - Draw vector $\vec{A}$ straight up $20$ km. - From the tip of $\vec{A}$, draw vector $\vec{B}$ at $60^\circ$ west of north with length $25$ km. - The resultant vector $\vec{R}$ is drawn from the start point to the end point. - Using scale and protractor, measure length and angle to verify algebraic results. This completes the solution.