1. Physics 201, Lect. 3 Jan. 26, 2010 Chapter 2: Motion in One Dimension

2. ©2008 by W.H. Freeman and Company Displacement: change in position “Displacement” only depends on the initial & final positions; “Distance” counts for the length of the full path.

3. ©2008 by W.H. Freeman and Company Distance and displacement of a dog (p. 29) total distance traveled = 35 ft net displacement = 5 ft

4. Average velocity and speed Velocity includes direction, and thus is a “vector”. It is fundamental for the laws of motion; speed is only of practical meaning.

5. ©2008 by W.H. Freeman and Company Average velocity and average speed of a dog (p 31) Average velocity = (net displacement)/(total time) = (+5 m)/(2.5 s) = +2 m/s Average speed = (total distance traveled)/(total time) = (35 m)/(2.5 s) = 14 m/s

6. The average velocity for a time interval (t 1, t 2 ) is the slope of the straight line connecting the points (t 1,x 1 ) and (t 2,x 2 ) on a graph of x versus t. ©2008 by W.H. Freeman and Company The average velocity of particle between times t 1 and t 2 ’ is greater than the average velocity of particle between times t 1 and t 2.

7. The instantaneous velocity along x, v x, is the limit of the ratio  x/  t as  t approaches zero.

8. ©2008 by W.H. Freeman and Company Approaching the limit  t  0 to find the instantaneous velocity

9. ©2008 by W.H. Freeman and Company Instantaneous velocity at a given time is the derivative of the x-versus-t curve. Although it is in units of m/s (displacement/time), independent of the distance/length.

10. ©2008 by W.H. Freeman and Company Displacement versus time when x = 5t 2, with x in meters and t in seconds (p 34). velocity = derivative of displacement dx/dt = 10 t m/s.

11. Acceleration Instantaneous acceleration along x: = derivative of v x -versus-t curve.

12. I nstantaneous acceleration along x is first derivative of v x with respect to time t, and second derivative of x with respect to t:

13. A skydiver jumps out A skydiver is falling straight down, along the negative y direction. During the initial part of the fall, her speed increases from 16 m/s to 28 m/s in 1.5 s. Which of the following is correct? y v a 1.v>0, a>0 2.v>0, a<0 3.v 0 4.v<0, a<0

14. A skydiver jumps out A skydiver is falling straight down, along the negative y direction. During the initial part of the fall, her speed increases from 16 m/s to 28 m/s in 1.5 s. Which of the following is correct? y v a 1.v>0, a>0 2.v>0, a<0 3.v 0 4.v<0, a<0 correct end 98

15. Parachute opens During a later part of the fall, after the parachute has opened, her speed decreases from 48 m/s to 26 m/s in 11 s. Which of the following is correct? y v a 1.v>0, a>0 2.v>0, a<0 3.v 0 4.v<0, a<0

16. Parachute opens During a later part of the fall, after the parachute has opened, her speed decreases from 48 m/s to 26 m/s in 11 s. Which of the following is correct? y v a 1.v>0, a>0 2.v>0, a<0 3.v 0 4.v<0, a<0 correct If speed is increasing, v and a are in same direction. If speed is decreasing, v and a are in opposite directions.

17. ©2008 by W.H. Freeman and Company Special case: constant acceleration If acceleration is constant, velocity increases linearly with time.

18. ©2008 by W.H. Freeman and Company Motion with constant acceleration When acceleration is constant, the average velocity is the velocity halfway through the time interval (THIS IS NOT ALWAYS TRUE IF THE ACCELERATION IS NOT CONSTANT!)

19. Motion with constant acceleration: find velocity and position versus time Consider a particle that at time t 0 has velocity along x of v x0 and is at position x 0. If the acceleration a x is constant, the velocity at time t is: v x (t) = v x0 + a x (t-t 0 ). See the area: v t  x. the particle’s position at time t, x(t), is x(t) = x 0 + v x0 (t-t 0 ) + (1/2)a x (t-t 0 ) 2. From the last figure,

20. Motion with constant acceleration: find velocity versus position Recall the particle’s velocity is: v x (t) = v x0 + a x (t-t 0 )  t-t 0 = (v x (t)-v x0 )/a x. Recall the particle’s position is x(t) = x 0 +v x0 (t-t 0 )+(1/2)a x (t-t 0 ) 2, this means x(t) = x 0 + v x0 (v x (t)-v x0 )/a x + (1/2)a x ((v x (t)-v x0 )/a x ) 2 or x(t)-x 0 = v x0 (v x (t)-v x0 )/a x + (1/2)(v x (t)-v x0 ) 2 /a x. Multiply both sides of equation by 2a x and simplifying yields 2a x (x(t)-x 0 ) = v x (t) 2 - v x0 2.

21. Motion with constant acceleration, simplified formulas: Setting the initial conditions x 0 = t 0 = 0, then at time t, v x = v x0 + a x t  t = ( v x -v x0 )/a x x = v x0 t + (1/2) a x t 2 Substitute t, 2 a x x = v x 2 – v x0 2

22. ©2008 by W.H. Freeman and Company Free fall: If no air, all objects in free-fall with same initial velocity move identically. acceleration due to gravity = g  9.81 m/s 2  32 ft/s 2, directed downwards (same for all objects) The acceleration results in a widening of the spaces between images. gravity demos