2. Physics 101: Lecture 4, Pg 1 Kinematics + Dynamics Physics 101: Lecture 04 l Today’s lecture will cover Textbook Chapter 4 Register late = excused absence. Visit undergraduate office 230/231 Loomis. If you are new to the course, please read the course description on the course web page (and email policy from Lecture 1 note)! neptune

3. Physics 101: Lecture 4, Pg 2 What’s Most Difficult I find vectors to be difficult to conceptualize. I understand forces are present but I am having trouble breaking them down into the x and y directions. Remembering all of my high school Trig. The class is moving a little too fast. I don't think the concepts are quite difficult yet, but I feel like the class is moving at fast pace.

4. Physics 101: Lecture 4, Pg 3 You can review vectors using the textbook!!! Pages 28-35 of Textbook This also includes trigonometry

5. Physics 101: Lecture 4, Pg 4 Review l Kinematics : Description of Motion è Position è Displacement  Velocity v =  x /  t »average »instantaneous  Acceleration a =  v /  t »average »instantaneous è Relative velocity: v ac = v ab + v bc

6. Physics 101: Lecture 4, Pg 5 Preflight 4.1 (A) (B)(C) Which x vs t plot shows positive acceleration? 89% got this correct!!!! “ The slope of the line is increasing with time, meaning that the speed is increasing. -Alexander R. ” …interpreting graphs… This is x(t) graph, not v(t)

7. Physics 101: Lecture 4, Pg 6 Equations for Constant Acceleration (text, page 113-114) l x = x 0 + v 0 t + 1/2 at 2 l v = v 0 + at l v 2 = v 0 2 + 2a(x-x 0 ) l  x = v 0 t + 1/2 at 2 l  v = at l v 2 = v 0 2 + 2a  x 14

8. Physics 101: Lecture 4, Pg 7 Kinematics Example l A car is traveling 30 m/s and applies its breaks to stop after a distance of 150 m. l How fast is the car going after it has traveled ½ the distance (75 meters) ? A) v 15 m/s

9. Physics 101: Lecture 4, Pg 8 Acceleration ACT A car accelerates uniformly from rest. If it travels a distance D in time t then how far will it travel in a time 2t ? A. D/4 B. D/2 C. D D. 2D E. 4D Correct x=1/2 at 2 Follow up question: If the car has speed v at time t then what is the speed at time 2t ? A. v/4 B. v/2 C. v D. 2v E. 4v Correct v=at Demo…

10. Physics 101: Lecture 4, Pg 9 Newton’s Second Law  F=ma position and velocity depend on history Net Force determines acceleration

11. Physics 101: Lecture 4, Pg 10 ACT l A force F acting on a mass m 1 results in an acceleration a 1.The same force acting on a different mass m 2 results in an acceleration a 2 = 2a 1. What is the mass m 2 ? (A) (B) (C) (A) 2m 1 (B) m 1 (C) 1/2 m 1 Fa1a1 m1m1 F a 2 = 2a 1 m2m2 F=ma F= m 1 a 1 = m 2 a 2 = m 2 (2a 1 ) Therefore, m 2 = m 1 /2 Or in words…twice the acceleration means half the mass

12. Physics 101: Lecture 4, Pg 11Example: y x X direction: Tractor  F = ma F Th – T = m tractor a F Th = T+ m tractor a T W N X direction: Trailer  F = ma T = m trailer a T F Th W N Combine: F Th = m trailer a + m tractor a F Th = (m trailer+ m tractor ) a A tractor T (m=300Kg) is pulling a trailer M (m=400Kg). It starts from rest and pulls with constant force such that there is a positive acceleration of 1.5 m/s 2. Calculate the horizontal thrust force on the tractor due to the ground. F Th = 1050 N

13. Physics 101: Lecture 4, Pg 12 Net Force ACT Compare F tractor the net force on the tractor, with F trailer the net force on the trailer from the previous problem. A) F tractor > F trailor B) F tractor = F trailor C) F tractor < F trailor  F = m a F tractor = m tractor a = (300 kg) (1.5 m/s 2 ) = 450 N F trailer = m trailer a = (400 kg) (1.5 m/s2) = 600 N 300Kg400Kg

14. Physics 101: Lecture 4, Pg 13 Pulley Example l Two boxes are connected by a string over a frictionless pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8 meters above the table, how long does it take to hit the table. 1 2 1)T - m 1 g = m 1 a 1 2)T – m 2 g = -m 2 a 1 2) T = m 2 g -m 2 a 1 1) m 2 g -m 2 a 1 - m 1 g = m 1 a 1 a 1 = (m 2 – m 1 )g / (m 1 +m 2 ) 1 T m1gm1g 2 T m2gm2g Compare the acceleration of boxes 1 and 2 A) |a 1 | > |a 2 | B) |a 1 | = |a 2 | C) |a 1 | < |a 2 | y x

15. Physics 101: Lecture 4, Pg 14 Pulley Example l Two boxes are connected by a string over a frictionless pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8 meters above the table, how long does it take to hit the table. 1 2 a 1 = (m 2 – m 1 )g / (m 1 +m 2 ) a = 2.45 m/s 2  x = v 0 t + ½ a t 2  x = ½ a t 2 t = sqrt(2  x/a) t = 0.81 seconds 1 T m1gm1g 2 T m2gm2g Compare the acceleration of boxes 1 and 2 A) |a 1 | > |a 2 | B) |a 1 | = |a 2 | C) |a 1 | < |a 2 | y x

16. Physics 101: Lecture 4, Pg 15 Summary of Concepts Constant Acceleration  x = x 0 + v 0 t + 1/2 at 2  v = v 0 + at  v 2 = v 0 2 + 2a(x-x 0 ) F = m a – Draw Free Body Diagram – Write down equations – Solve Next time: textbook section 4.3, 4.5