Motion in Two Dimensions

1. Motion in Two Dimensions Chapter 4 (Part 1)

2. Vectors Vectors have magnitude (length) and direction In previous chapters, we have treated “direction” as being only positive or negative Vectors are represented by arrows; length of the arrow corresponds to magnitude (length) of the vector

3. Vectors Vector quantities include displacement, velocity, and acceleration Scalar quantities (opposite of vectors) have only one value Examples include: distance & speed

4. Resolving Vectors

5. Resolving Vectors Displacement Vector

6. Resolving Vectors Displacement Vector 950 m 440 m

7. Resolving Vectors 1047 m NE 950 m 440 m

8. Resolving Vectors (using Trig) An ant leaves his nest and, after foraging for some time, is at the location given by the vector r in the diagram below. This vector has the magnitude r = 1.50 m and Θ = 25.0°. Resolve this vector into its components using sine and cosine. r = 1.50 m r y r x Θ = 25.0°

9. Resolving Vectors (using Trig) An ant leaves his nest and, after foraging for some time, is at the location given by the vector r in the diagram below. This vector has the magnitude r = 1.50 m and Θ = 25.0°. Resolve this vector into its components using sine and cosine. r = 1.50 m r y r x Θ = 25.0°  

10. Resolving Vectors (using Trig) An ant leaves his nest and, after foraging for some time, is at the location given by the vector r in the diagram below. This vector has the magnitude r = 1.50 m and Θ = 25.0°. Resolve this vector into its components using sine and cosine. r = 1.50 m r y r x Θ = 25.0° Rearrange this equation…  

11. Resolving Vectors (using Trig) An ant leaves his nest and, after foraging for some time, is at the location given by the vector r in the diagram below. This vector has the magnitude r = 1.50 m and Θ = 25.0°. Resolve this vector into its components using sine and cosine. r = 1.50 m r y r x Θ = 25.0° Rearrange this equation…  

12. Resolving Vectors (using Trig) An ant leaves his nest and, after foraging for some time, is at the location given by the vector r in the diagram below. This vector has the magnitude r = 1.50 m and Θ = 25.0°. Resolve this vector into its components using sine and cosine. r = 1.50 m r y r x Θ = 25.0°  

13. Resolving Vectors (using Trig) An ant leaves his nest and, after foraging for some time, is at the location given by the vector r in the diagram below. This vector has the magnitude r = 1.50 m and Θ = 25.0°. Resolve this vector into its components using sine and cosine. r = 1.50 m r y r x Θ = 25.0° Rearrange this equation…  

14. Resolving Vectors (using Trig) An ant leaves his nest and, after foraging for some time, is at the location given by the vector r in the diagram below. This vector has the magnitude r = 1.50 m and Θ = 25.0°. Resolve this vector into its components using sine and cosine. r = 1.50 m r y r x Θ = 25.0° Rearrange this equation…  

15. Combining Vectors

16. Combining Vectors 0.60 km 0.60 km ? To find magnitude, use Pythagorean theorem.  

17. Combining Vectors 0.60 km 0.60 km ? To find magnitude, use Pythagorean theorem. 0.72 m 2 = c 2 c = 0.85 m  

18. Combining Vectors 0.60 km 0.60 km ? To find direction, use trigonometry  

19. Combining Vectors 0.60 km 0.60 km ? To find direction, use trigonometry  

20. Practice Problems Resolve this vector into its horizontal (x) and vertical (y) components: A cannonball is fired with a velocity of 34.0 m/s at an angle of 31.5° to the horizontal. Combine these two vectors: A plan flies due west at 245 m/s and is pushed by a northward wind blowing at 29.0 m/s. What is the resultant vector for the plane’s velocity? Hint #1: Draw a picture for each! Hint #2: To determine sides of a triangle use trig functions. To determine an angle, use inverse trig functions (tan -1 , etc.)

21. Challenge Problem Resolve each of the three vectors to answer the following question. A man is exploring with a compass in the dark. He is 300 meters west from a lake, the coastline of which extends due north and south. He follows the following three directions. (All angles are measured from the horizontal.) Walk 250 meters NE at 40 degrees. Walk 100 m NW at 30 degrees. Walk 300 m NE at 75 degrees. How far is he from the lake? Lake 1 2 3 ? 300 m

22. Challenge Problem Combine the three vectors to answer the following question. A man is exploring with a compass in the dark. He is 300 meters west from a lake, the coastline of which extends due north and south. He follows the following three directions. (All angles are measured from the horizontal. Walk 250 meters NE at 40 degrees. Walk 100 m NW at 30 degrees. Walk 300 m NE at 75 degrees. What is his resultant vector? Lake 1 2 3 ?

23. Relative motion Chapter 4 (Part 2)

24. Relative Motion Often, motion must be described relative to the motion of another object. We do this all the time without realizing it… for instance, your car drives 50 mi/ hr down a highway relative to the motion of the ground. Think about it: A train is moving at 15.0 m/s. A man walks towards the front of the train at 1.2 m/s. What is his speed relative to the ground? What if he is walking at 1.2 m/s towards the back of the train?

25. Relative Motion If the motion occurs in two dimensions, simply add the vectors using the strategies from earlier in the chapter.

26. Practice Problem

27. Projectile motion Chapter 4 (Part 3)

28. Projectile Motion An object that is thrown will follow a curved path (parabola) to an onlooker. Air resistance must be ignored (so horizontal velocity is constant) Acceleration due to gravity is -9.81 m/s 2 .

29. Position-Time Equations for Projectiles Horizontal motion occurs at constant velocity… Vertical motion includes downward acceleration due to gravity…  

30. Practice Problem

31. Practice Problems You throw a baseball off the roof of a house at 25.0 m/s, perfectly parallel to the ground below. How far has it traveled towards the ground after 2.00 seconds? How far has it traveled away from you (horizontally) after 2.00 seconds?

32. Practice Problem Yarrrr ! An angry pirate launches a cannonball at a ship over yonder. The cannon is initially set at an angle of 25.0° and, with enough gunpowder, can launch the cannonball at an initial speed of 42.0 m/s along that angle. The deck of the ship (with the cannon on it) is 8.00 meters above the water. How much time will pass before the cannonball lands in the water below? How far away from the ship will the cannonball be when it lands?