1. Motion in Two and Three Dimensions Chapter 4

2. Position and Displacement A position vector locates a particle in space o Extends from a reference point (origin) to the particle Example o Position vector (-3m, 2m, 5m) Eq. (4-1) Figure 4-1

3. Position and Displacement Change in position vector is a displacement We can rewrite this as: Or express it in terms of changes in each coordinate: Eq. (4-2) Eq. (4-3) Eq. (4-4)

4. Average Velocity and Instantaneous Velocity Average velocity is o A displacement divided by its time interval We can write this in component form: Example o A particle moves through displacement (12 m)i + (3.0 m)k in 2.0 s: Eq. (4-8) Eq. (4-9)

5. Average Velocity and Instantaneous Velocity Instantaneous velocity is o The velocity of a particle at a single point in time o The limit of avg. velocity as the time interval shrinks to 0 Visualize displacement and instantaneous velocity: Eq. (4-10) Figure 4-4 Figure 4-3

6. Average Velocity and Instantaneous Velocity In unit-vector form, we write: Which can also be written: Note: a velocity vector does not extend from one point to another, only shows direction and magnitude Eq. (4-11) Eq. (4-12)

7. Average Acceleration and Instantaneous Acceleration Average acceleration is o A change in velocity divided by its time interval Instantaneous acceleration is again the limit t → 0: We can write Eq. 4-16 in unit-vector form: Eq. (4-15) Eq. (4-16)

8. Average Acceleration and Instantaneous Acceleration We can rewrite as: To get the components of acceleration, we differentiate the components of velocity with respect to time Eq. (4-17) Eq. (4-18) Figure 4-6

9. Average Acceleration and Instantaneous Acceleration © 2014 John Wiley & Sons, Inc. All rights reserved.

10. Average Acceleration and Instantaneous Acceleration © 2014 John Wiley & Sons, Inc. All rights reserved.

11. Average Acceleration and Instantaneous Acceleration HW#: At t = 0 the velocity of a particle is measured to be 5 m/s along the positive x–axis and its acceleration is a = -3î + 4ĵ (m/s 2 ), Find The velocity vector as a function of time ? {answer v = (5 – 3t)î + 4t ĵ} © 2014 John Wiley & Sons, Inc. All rights reserved.

12. Characteristics of Projectile Motion: Horizontal Motion of a ball rolling freely along a level surface. Horizontal velocity is always constant. Vertical Motion of a freely falling object. Force due to gravity. Vertical component of velocity changes with time. Projectile Motion

13. A projectile is o A particle moving in the x-y plane o With some initial velocity o Whose acceleration is always free-fall acceleration (g) The motion of a projectile is projectile motion Launched with an initial velocity v 0 Eq. (1) Eq. (2)

14. Projectile Motion Horizontal motion: o No acceleration, so velocity is constant ( a x =0 ) : Vertical motion: o Acceleration is always g ( a y = g ): Eq. (3) Eq. (4) Eq. (5) Eq. (6)

15. Projectile Motion Horizontal range and the maximum height of a projectile: To find the maximum height we use the equation H = v 2 o sin 2 θ / 2g © 2014 John Wiley & Sons, Inc. All rights reserved.

16. Projectile Motion

17. EXAMPLE : A ball is thrown with an initial velocity of 20 m/s at an angle of 30 above the ground. Find The maximum height the ball can reach? The range of the ball travel? The time taken to hit the ground again? © 2014 John Wiley & Sons, Inc. All rights reserved.

18. Projectile Motion Solve: H = v 2 o sin 2 θ / 2g = (20 x sin 30) 2 / (2 x 9.8 ) = 5.1 m R = v 2 o sin2θ / g = (400 x sin 60) / 9.8 = 35.3 m. v y = v o sinθ – gt v y (at maximum height) = 0 t = v o sinθ / g = 20 x sinθ / 9.8 t = 1.02 s (time needed to achieve the maximum height) T totall = 2 x t = 2.04 s © 2014 John Wiley & Sons, Inc. All rights reserved.

19. Projectile Motion EXAMPLE: A ball is thrown horizontally with an initial velocity of 25 m/s from the top of a 60 m high building. How much time will the ball take to hit the ground ? How far does the ball travel horizontally ? What is the velocity of the ball as it hits the ground ? © 2014 John Wiley & Sons, Inc. All rights reserved.

20. Projectile Motion solve y = y 0 + (v 0 sin ѳ )t – ½ gt 2 Ѳ = 0 y = -1/2gt 2 t = 3.5s x = (v 0 cos ѳ )t x = 87.5 m v x = v 0 cos Ѳ =25 m/s v y = v 0 sin Ѳ – gt = -34.3 m/s | v |= 42.44 m/s © 2014 John Wiley & Sons, Inc. All rights reserved.

21. Projectile Motion Example : : a projectile is fired to achieve a maximum range of 120 m.what should its speed be ? R max =V 2 0 /g V 0 =37 m/s Example 4 : a projectile is fired with an angle Θ above the horizontal. take 15s to reach its range of 110 m. What is its speed at the highest point ? x = (v 0 cos Ѳ )t v 0 cos Ѳ = v 0x = x/t = 9.34 m/s © 2014 John Wiley & Sons, Inc. All rights reserved.

22. Projectile Motion HW #: A particle is projected with an initial velocity v i = ( 10i ̂ + 10j ̂ ) m/s The angle of projection is : ( answer = 45 ֯ ) © 2014 John Wiley & Sons, Inc. All rights reserved.

23. Uniform Circular Motion A particle is in uniform circular motion if It travels around a circle path The tangential velocity of a particle has a constant magnitude and change direction continuously The centripetal acceleration has a constant magnitude and towards at the center of the circle The velocity and the acceleration perpendicular to each other Figure 4-16

24. Uniform Circular Motion To Find centripetal acceleration use the equation ɑ - the centripetal acceleration ν - the velocity of the object r - the radius of the circle The period of revolution is: The time it takes to complete one revolution The Frequency the number of revolution per second

25. Uniform Circular Motion

28. Position Vector Locates a particle in 3-space Displacement Change in position vector Average and Instantaneous Accel. Average and Instantaneous Velocity Eq. (4-2) Eq. (4-8) Eq. (4-15) Summary Eq. (4-1) Eq. (4-3) Eq. (4-4) Eq. (4-10) Eq. (4-16)

29. Projectile Motion Flight of particle subject only to free-fall acceleration (g) Uniform Circular Motion Magnitude of acceleration: Time to complete a circle: Eq. (4-34) Summary Eq. (4-22) Eq. (4-23) Eq. (4-35)

30. EXERCISE Exercise 1, 3, 6, 13, 56, 62 © 2014 John Wiley & Sons, Inc. All rights reserved.