Physics 1710 Chapter 2 Motion in One Dimension—II

1. Physics 1710Chapter 2 Motion in One Dimension—II Galileo Galilei Linceo (1564-1642) Discourses and Mathematical Demonstrations Concerning the Two New Sciences (1638) at age 74! (four years before his death at 78)

2. Physics 1710Chapter 2 Motion in One Dimension Galileo’s Ramp

3. Physics 1710Chapter 2 Motion in One Dimension—II 1′ Lecture Under uniform acceleration ax: Acceleration is constant (hence “uniform”):ax = constant (>0,=0, or <0) Velocity changes linearly in time: vfinal = vinitial + ax t Displacement increases quadratically with time:xfinal = xinitial+vinitialt + ½ ax t 2

4. Physics 1710Chapter 2 Motion in One Dimension—II Displacement is the change in position. Δx =xfinal - xinitial Δ is the change “operator”: Change in x =x at end– xat start ΔBalance =Balancefinal - Balanceinitial REVIEW

5. Physics 1710Chapter 2 Motion in One Dimension—II Velocity is the time rate of displacement. Average velocity: vx, ave = Δ x / Δt Instantaneous velocity: vx = lim Δt→∞Δx /Δt = dx/dt REVIEW

6. Δx Δt Physics 1710Chapter 2 Motion in One Dimension—II Position (m) vaverage =Δx/Δt Time (sec)

7. Δx Δt Physics 1710Chapter 2 Motion in One Dimension—II Plot it! vx = dx/dt Instantaneous Velocity Position (m) vx, ave = Δx / Δt AverageVelocity Velocity Time (sec)

8. Physics 1710Chapter 2 Motion in One Dimension—II Acceleration is the time rate of change of velocity. Average acceleration: ax, ave = Δvx / Δt = (vx,final -vx, initial )/ Δt Instantaneous acceleration: ax = lim Δt→∞Δvx / Δt = dvx /dt ax = dvx /dt =d(dxx /dt)/dt = d2x/dt 2

9. Physics 1710Chapter 2 Motion in One Dimension—II Motion Map From “snap shots” of motion at equal intervals of time we can determine the displacement, the average velocity and the average acceleration in each case. Uniform motion Accelerated motion

10. vx = dx/dt Physics 1710Chapter 2 Motion in One Dimension Plot them! a2>0 Velocity a1= 0 Position (m) Velocity (m/sec) ax = dvx /dt Time (sec) Time (sec)

11. Physics 1710Chapter 2 Motion in One Dimension—II Galileo Galilei Linceo

12. Physics 1710Chapter 2 Motion in One Dimension—II Galileo’s rule of odd numbers: Under uniform acceleration from rest, a body will traverse distances in successive equalintervals of time that stand in ratio as the odd numbers 1,3,5,7,9 …

13. Physics 1710Chapter 2 Motion in One Dimension—II Galileo’s Ramp Demonstration

14. Physics 1710Chapter 2 Motion in One Dimension—II Observation: 0 : 0 = 0 2 : 0+1 =1 = 1 2 : 1+3 = 4 = 2 2 : 4+5 = 9 = 3 2 : 9+7= 16 = 4 2 : 16+9= 25 = 5 2 : 25+11= 36 = 6 2 ∆x ∝ t 2 ; from rest (vinitial = 0)

15. Physics 1710Chapter 2 Motion in One Dimension—II (dv/dt) = a = constant ∫0t (dv/dt) dt = ∫0t a dt ∫vinitialvfinal dv = a (t-0) ∆v = v– vinitial = at The change in the instantaneous velocity is equal to the (constant) acceleration multiplied by its duration. v= vinitial + at Kinematic Equations from Calculus:

16. Physics 1710Chapter 2 Motion in One Dimension—II dx/dt = v ∫0t (dx/dt) dt = ∫0t vdt x– xinitial = ∫0t (vinitial + at) dt = vinitial (t-0) + ½ a(t 2-0) ∆x = vinitial t + ½ at 2 The displacement under uniform acceleration is equal to the displacement at constant velocity plus one half the acceleration multiplied by the square of its duration. x= xinitial + vinitial t + ½ at 2 Kinematic Equations from Calculus:

17. Physics 1710Chapter 2 Motion in One Dimension—II v 2 = (vinitial + at) 2 v 2 = vinitial 2 +2 vinitial at + a 2 t 2 v 2 = vinitial 2 + 2a (vinitial t + ½ at 2) v 2 = vinitial 2 +2a∆x The change in the square of the velocity is equal to two times the acceleration multiplied by the distance over which the acceleration is applied. Kinematic Equations from Calculus:

18. Physics 1710Chapter 2 Motion in One Dimension—II 80/20 facts: Kinematic Equations (1-D Uniform a): vx final = vx initial + ax t x final = x initial + vx initial t + ½ axt 2 vx final2 = vx initial2 + 2 ax (x final - x initial )

19. Physics 1710Chapter 2 Motion in One Dimension—II 80/20 facts: In free fall near the Earth, all bodies are accelerated uniformly downward with an acceleration of az = - g = -9.80 m/s2.

20. Physics 1710Chapter 2 Motion in One Dimension—II Plot it! Position (m) Acceleration (m/s/s) Velocity (m/sec) 0 0 dx/dt → dv/dt → -9.8 m/s/s Time (sec) Time (sec) Time (sec)

21. The change in the instantaneous velocity is equal to the (constant) acceleration multiplied by its duration. ∆v = at • The displacement is equal to the displacement at constant velocity plus one half of the product of the acceleration and the square of its duration. ∆x = vinitial t + ½ at 2 • The change in the square of the velocity is equal to two times the acceleration multiplied by the distance traveled during acceleration. ∆v 2 = 2a ∆x • The acceleration of falling bodies is 9.8 m/s/s downward.a = - g = - 9.8 m/s/s Physics 1710Chapter 2 Motion in One Dimension—II Summary:

22. Physics 1710Chapter 2 Motion in One Dimension—II • The main point of today’s lecture. • A realization I had today. • A question I have. 1′ Essay:One of the following: