1. Physics 101: Lecture 3, Pg 1 Kinematics Physics 101: Lecture 03 l Today’s lecture will cover Textbook Sections 3.1-3.3

2. Physics 101: Lecture 3, Pg 2 Force at Angle Example A person is pushing a 15 kg block across a floor with  k = 0.4 at a constant speed. If she is pushing down at an angle of 25 degrees, what is the magnitude of her force on the block? y x Normal Weight Pushing x- direction:  F x = ma x F push cos(  ) – F friction = 0 F push cos(  ) –  F Normal = 0 F Normal = F push cos(  ) /  y- direction:  F y = ma y F Normal –F weight – F Push sin(  ) = 0 F Normal –mg – F Push sin(  ) = 0 Combine: F push cos(  ) /  –mg – F Push sin(  ) = 0 F push ( cos(  ) /  - sin(  )) = mg F push = m g / ( cos(  )/  – sin(  ))  Friction  09 F push = 80 N

3. Physics 101: Lecture 3, Pg 3 Force at Angle Example A person is pushing a 15 kg block across a floor with  k = 0.4 at a constant speed. If she is pushing down at an angle of 25 degrees, what is the magnitude of her force on the block? y x  09

4. Physics 101: Lecture 3, Pg 4 Homework Example l Calculate the tension in the left string. x-direction:  F=ma -T L +T R cos(  )= 0 T L = T R cos(  ) y-direction:  F=ma T R sin(  ) – Mg = 0 T R = Mg / sin(  ) Combine: T L = Mg cos(  )/sin(  ) TLTL W  TRTR 07 y x

5. Physics 101: Lecture 3, Pg 5 Homework Example l Calculate the tension in the left string. 07 y x A) Do it B) Skip it

6. Physics 101: Lecture 3, Pg 6 Overview l Kinematics: Description of Motion è Position and displacement è velocity »average »instantaneous è acceleration »average »instantaneous 10

7. Physics 101: Lecture 3, Pg 7 Position vs Time Plots l Gives location at any time. l Displacement is change in position. l Slope gives velocity. 12 x (m) t 4 3 -3 Position at t=3, x(3) = Displacement between t=5 and t=1.  x = Average velocity between t=5 and t=1. v = 1 -1.0 m -0.25 m/s 1.0 m - 2.0 m = -1.0 m -1 m / 4 s = -0.25 m/s

8. Physics 101: Lecture 3, Pg 8 1.5 6 Velocity vs Time Plots l Gives velocity at any time. l Area gives displacement l Slope gives acceleration. 18 v (m/s) t 4 3 -3 velocity at t=2, v(2) = Change in v between t=5 and t=3.  v = Average acceleration between t=5 and t=3: a = Displacement between t=0 and t=3:  x= Average velocity between t=0 and t=3? v= 3 m/s 7.5 m t=0 to t=1: ½ (3m/s) (1 s) = 1.5 m 7.5 m / 3s = 2.5 m/s -2 m/s – 3 m/s = -5 m/s -5 m/s / (2 s) = -2.5 m/s 2 t=1 to t=3: (3m/s) (2 s) = 6 m

9. Physics 101: Lecture 3, Pg 9 2 6 Acceleration vs Time Plots l Gives acceleration at any time. l Area gives change in velocity 21 a (m/s 2 ) t 4 3 -3 Acceleration at t=4, a(4) = Change in v between t=4 and t=1.  v = -2 m/s 2 t=1-3:  v = (3m/s 2 )(2s) = 6 m/s t=3-4:  v = (-2m/s 2 )(1s) = -2 m/s +4 m/s

10. Physics 101: Lecture 3, Pg 10 Is it possible for an object to have a positive velocity at the same time as it has a negative acceleration? 1 - Yes 2 - No Acceleration Preflights 24 If the velocity of some object is not zero, can its acceleration ever be zero ? 1 - Yes 2 - No 83% 17% 86% 14% “It is possible for something to be moving forward (positive velocity) but be slowing down (declerating)” “If you have constant velocity (a car going at 55 mph) then your acceleration would be zero.” “An example of this is a ball thrown up in the air…

11. Physics 101: Lecture 3, Pg 11 Velocity ACT If the average velocity of a car during a trip along a straight road is positive, is it possible for the instantaneous velocity at some time during the trip to be negative? A - Yes B - No 27 Drive north 5 miles, put car in reverse and drive south 2 miles. Average velocity is positive.

12. Physics 101: Lecture 3, Pg 12 Dropped Ball l Draw v y vs t A ball is dropped from a height of two meters above the ground. 34 -6 v t 0.5 9 -6 v t 0.5 9 -6 v t 0.5 9 -6 v t 0.5 9 v t 9 -6 AB C D E y x

13. Physics 101: Lecture 3, Pg 13 Dropped Ball l Draw v vs t l Draw x vs t l Draw a vs t v t x t a t A ball is dropped for a height of two meters above the ground. 36

14. Physics 101: Lecture 3, Pg 14 Tossed Ball l Draw v vs t A ball is tossed from the ground up a height of two meters above the ground. And falls back down 42 vvv -6 t 1 9 t 1 9 t 1 9 v t 1 9 v t 1 9 ABC DE y x

15. Physics 101: Lecture 3, Pg 15 Tossed Ball l Draw v vs t l Draw x vs t l Draw a vs t v t x t a t 45 A ball is tossed from the ground up a height of two meters above the ground. And falls back down

16. Physics 101: Lecture 3, Pg 16 Not 0 since V f and V i are not the same ! = 0 A ball is thrown straight up in the air and returns to its initial position. During the time the ball is in the air, which of the following statements is true? A - Both average acceleration and average velocity are zero. B - Average acceleration is zero but average velocity is not zero. C - Average velocity is zero but average acceleration is not zero. D - Neither average acceleration nor average velocity are zero. ACT V ave =  Y/  t = (Y f – Y i ) / (t f – t i ) a ave =  V/  t = (V f – V i ) / (t f – t i ) 48

17. Physics 101: Lecture 3, Pg 17 Where is velocity zero? Where is velocity positive? Where is velocity negative? Where is speed largest? Where is acceleration zero? Where is acceleration positive? position vs. time velocity vs. time Example 48

18. Physics 101: Lecture 3, Pg 18 Summary of Concepts l kinematics: A description of motion l position: your coordinates l displacement:  x = change of position l velocity: rate of change of position è average :  x/  t è instantaneous: slope of x vs. t l acceleration: rate of change of velocity è average:  v/  t è instantaneous: slope of v vs. t 50

19. Physics 101: Lecture 3, Pg 19 Acceleration (m/s 2 ) l The average acceleration is the change in velocity divided by the change in time. Instantaneous acceleration is limit of average velocity as  t gets small. It is the slope of the v(t) plot t v(t) tt vv t 31

20. Physics 101: Lecture 3, Pg 20 Velocity l Slope of x vs t gives v vs t Area under v vs t gives  x! l Which plot best represents v(t) x(t) tt xx v(t) t t tt 5s 4s