Subject: General Physics 1
Lesson Vector Addition
Show your answers and solutions in each question
1. Jason walks 3.0 m east, stops to rest, and runs 9.0 m
north. What is his resultant displacement?
2. Mimi walks home from school 85 m west and remembers
she has to bring home her science book which a
classmate borrowed. She then turns 102 m south to her
classmate’s house. What is her resultant displacement?
3. A boat heading 7.20 km/h west crosses a river which has
a current drag of 11.0 km/h north. What is the resultant
velocity of the boat?
4. Two tugboats pull on a ship. Tugboat A pulls with a
velocity of 6.40 m/s south. Tugboat B pulls at 9.30 m/s
east. What is the ship’s resultant velocity?
5. A car heads 32 m south, then 41 m east, and then 55 m
north. What is the resultant displacement?
Answer (1)
Here are the solutions for each vector addition problem: 1. Jason's Displacement - Diagram: Imagine a right triangle. Jason's eastward walk is the horizontal leg (3.0 m), and his northward run is the vertical leg (9.0 m). The hypotenuse represents his resultant displacement.- Calculations: - We use the Pythagorean theorem: Resultant displacement² = 3² + 9² = 90- Resultant displacement = √90 ≈ 9.49 m- We need to find the angle (θ) of the resultant displacement using trigonometry: tan(θ) = opposite/adjacent = 9/3 = 3- θ = tan⁻¹(3) ≈ 71.6°- Answer: Jason's resultant displacement is approximately 9.49 meters at 71.6° north of east. 2. Mimi's Displacement - Diagram: Similar to the previous problem, imagine a right triangle. Mimi's westward walk is the horizontal leg (85 m), and her southward walk is the vertical leg (102 m).- Calculations: - Resultant displacement² = 85² + 102² = 17389- Resultant displacement = √17389 ≈ 131.8 m- θ = tan⁻¹(102/85) ≈ 50.3°- Answer: Mimi's resultant displacement is approximately 131.8 meters at 50.3° south of west. 3. Boat's Resultant Velocity - Diagram: Imagine a right triangle. The boat's westward velocity is the horizontal leg (7.20 km/h), and the river current's northward velocity is the vertical leg (11.0 km/h).- Calculations: - Resultant velocity² = 7.2² + 11² = 185.84- Resultant velocity = √185.84 ≈ 13.63 km/h- θ = tan⁻¹(11/7.2) ≈ 56.8°- Answer: The boat's resultant velocity is approximately 13.63 km/h at 56.8° north of west. 4. Ship's Resultant Velocity - Diagram: This is a vector addition problem where the vectors are not perpendicular. We'll use the law of cosines and the law of sines to solve for the resultant velocity.- Calculations: - The angle between the two tugboats' velocities is 90° (south and east are perpendicular).- Using the law of cosines: Resultant velocity² = 6.4² + 9.3² - 2(6.4)(9.3)cos(90°)- Resultant velocity² ≈ 121.33- Resultant velocity ≈ 11.01 m/s- Using the law of sines to find the angle: sin(θ)/9.3 = sin(90°)/11.01- θ ≈ 57.7°- Answer: The ship's resultant velocity is approximately 11.01 m/s at 57.7° south of east. 5. Car's Resultant Displacement - Diagram: This problem involves three displacements. It's helpful to visualize the car's movements on a coordinate plane. The car travels south (negative y-direction), then east (positive x-direction), and finally north (positive y-direction).- Calculations: - Break down the displacements into x and y components:- Southward displacement: (0, -32)- Eastward displacement: (41, 0)- Northward displacement: (0, 55)- Add the x and y components separately:- Total x displacement: 0 + 41 + 0 = 41 m- Total y displacement: -32 + 0 + 55 = 23 m- Calculate the resultant displacement using the Pythagorean theorem: Resultant displacement² = 41² + 23² = 2178- Resultant displacement = √2178 ≈ 46.67 m- θ = tan⁻¹(23/41) ≈ 29.3°- Answer: The car's
