How to calculate the resultant vector from 10 km/hr and 2 km/hr and determine its direction?

Steps to Solve

1. Identify the vectors We have two vectors: one with a magnitude of $10 \text{ km/hr}$ (car) and the other with $2 \text{ km/hr}$ (wind). We will represent the car's velocity horizontally and the wind's velocity vertically.

2. Draw the vectors head-to-tail Position the car's vector horizontally to the right and the wind's vector vertically upwards, forming a right triangle with these two vectors.

3. Apply the Pythagorean theorem To find the magnitude of the resultant vector $R$, use the Pythagorean theorem: $$ R = \sqrt{(10^2) + (2^2)} $$

4. Calculate the magnitude Substituting the values into the equation: $$ R = \sqrt{(10^2) + (2^2)} = \sqrt{100 + 4} = \sqrt{104} $$ Calculating $\sqrt{104}$ gives approximately: $$ R \approx 10.198 \text{ km/hr} $$

5. Determine the direction To find the angle $\theta$, use the tangent function: $$ \tan \theta = \frac{2}{10} $$ Then, solve for $\theta$: $$ \theta = \tan^{-1}\left(\frac{2}{10}\right) $$ Calculating this yields: $$ \theta \approx 11.31^\circ $$

6. State the final resultant vector The resultant vector is: $$ R \approx 10.198 \text{ km/hr}, \text{ Direction: } E11.31^\circ N $$