Explain: Certainly! In scenario 6c, you have velocities of 15 km/h due north and 23 km/h on a bearing of 253°. To find the resultant velocity, you can use vector addition. The two velocities form a right-angled triangle, where the northward velocity is one leg, the eastward velocity is the other leg, and the resultant velocity is the hypotenuse.
1. First, break down the 23 km/h velocity into its northward and eastward components using trigonometry. Since the angle is given as 253°, the eastward component is \(23 \cdot \cos(253°)\) and the northward component is \(23 \cdot \sin(253°)\).
2. Add the northward component of the 23 km/h velocity to the 15 km/h northward velocity. This gives you the total northward component.
3. Add the eastward component of the 23 km/h velocity to zero (since there's no eastward component in the 15 km/h northward velocity). This gives you the total eastward component.
4. Use these total northward and eastward components to find the magnitude and direction of the resultant velocity using the Pythagorean theorem and trigonometry.
calculations: I'll guide you through the calculations for 6c:
1. Eastward component of the 23 km/h velocity: \(23 \cdot \cos(253°) \approx -20.6\) km/h (negative because it's in the westward direction).
2. Northward component of the 23 km/h velocity: \(23 \cdot \sin(253°) \approx -9.4\) km/h (negative because it's in the southward direction).
Now, add these components to the initial velocity:
- Total northward velocity: \(15 - 9.4\) km/h.
- Total eastward velocity: \(-20.6\) km/h.
Use the Pythagorean theorem to find the magnitude of the resultant velocity:
\[ |R| = \sqrt{(\text{Total Northward Velocity})^2 + (\text{Total Eastward Velocity})^2} \]
Once you find \(|R|\), use trigonometry to find the direction:
\[ \theta = \arctan\left(\frac{\text{Total Eastward Velocity}}{\text{Total Northward Velocity}}\right) \]
