Like this presentation? Why not share!

- 5.7 notes by Mrs. Hedrick's Class 65 views
- 20 July Presentation by DonBrown3PM 305 views
- Survey results by reuben95 70 views
- Unique Hoodia - Are you Prepared Fo... by winsylee2020 118 views
- Socio psychological anslysis of ge... by payal 340 views
- Sponsors of the Tweet to Be Heard C... by William Michael 272 views

651

-1

-1

Published on

No Downloads

Total Views

651

On Slideshare

0

From Embeds

0

Number of Embeds

0

Shares

0

Downloads

44

Comments

0

Likes

2

No embeds

No notes for slide

- 1. Exam IPhysics 101: Lecture 03 Kinematics Today’s lecture will cover Textbook Sections 3.1-3.3 (and some Ch. 4) Physics 101: Lecture 3, Pg 1
- 2. AnnouncementsHW 1 is due Tuesday Aug 31 (tomorrow) at 6am.No clicker points for Lect. 2? Re-register.Office hours now underway.Read the course description & FAQ on the course web site! Physics 101: Lecture 3, Pg 2
- 3. Force at Angle Example A person is pushing a 15 kg block across a floor with mk= 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?x- direction: SFx = max Combine: Fpush cos(q) – Ffriction = 0 Fpush cos(q) / m–mg – FPush sin(q) = 0 Fpush cos(q) – m FNormal = 0 Fpush ( cos(q) / m - sin(q)) = mg FNormal = Fpush cos(q) / m Fpush = m g / ( cos(q)/m – sin(q))y- direction: SFy = may Fpush = 80 NFNormal –Fweight – FPush sin(q) = 0 NormalFNormal –mg – FPush sin(q) = 0 Pushing y q x q Friction Weight Physics 101: Lecture 3, Pg 3
- 4. Homework 2 Example Calculate the tension in the left string. Work through example? y A: Yes B: No x x-direction: SF=ma -TL+TR cos(q)= 0 TR TL q TL = TR cos(q) y-direction: SF=ma W TR sin(q) – Mg = 0 Combine: TR = Mg / sin(q) TL = Mg cos(q)/sin(q) Physics 101: Lecture 3, Pg 4
- 5. Overview Kinematics: Description of Motion Position and displacement velocity » average » instantaneous Acceleration » average » instantaneous Relative velocity (first pass) Physics 101: Lecture 3, Pg 5
- 6. Position vs Time Plots Gives location at any time. Displacement is change in position. Slope gives velocity. x (m) 3Position at t=3, x(3) = 1Displacement between t=5 and t=1. Dx = -1.0 m t 4 1.0 m - 2.0 m = -1.0 mAverage velocity between t=5 and t=1. v = -0.25 m/s -3 -1 m / 4 s = -0.25 m/s Physics 101: Lecture 3, Pg 6
- 7. Velocity vs Time Plots v (m/s) Gives velocity at any time. Area gives displacement 3 Slope gives acceleration. 1.5 6 t velocity at t=2, v(2) = 3 m/s 4 Displacement between t=0 and t=3: Dx= 7.5 m -3 t=0 to t=1: ½ (3m/s) (1 s) = 1.5 m t=1 to t=3: (3m/s) (2 s) = 6 mAverage velocity between t=0 and t=3? v= 7.5 m / 3s = 2.5 m/sChange in v between t=5 and t=3. Dv = -2 m/s – 3 m/s = -5 m/sAverage acceleration between t=5 and t=3: a = -5 m/s / (2 s) = -2.5 m/s2 Physics 101: Lecture 3, Pg 7
- 8. Acceleration vs Time Plots Gives acceleration at any time. Area gives change in velocity a (m/s2) 3Acceleration at t=4, a(4) = -2 m/s2 6Change in v between t=4 and t=1. Dv = +4 m/s t t=1-3: Dv = (3m/s2)(2s) = 6 m/s 24 t=3-4: Dv = (-2m/s2)(1s) = -2 m/s -3 Physics 101: Lecture 3, Pg 8
- 9. Acceleration Preflights Is it possible for an object to have a positive velocity at the same time as it has a negative acceleration?88% 1 - Yes “the object could be slowing down.”12% 2 - No If the velocity of some object is not zero, can its acceleration ever be zero ?87% 1 - Yes “The velocity could be non-zero and constant. A constant velocity has no13% 2 - No acceleration..” Physics 101: Lecture 3, Pg 9
- 10. Velocity ACTIf the average velocity of a car during a trip along a straightroad is positive, is it possible for the instantaneous velocity atsome time during the trip to be negative?A - YesB - NoDrive north 5 miles, put car in reverseand drive south 2 miles. Averagevelocity is positive. Physics 101: Lecture 3, Pg 10
- 11. Dropped Ball y•A ball is dropped from a height of xtwo meters above the ground. Draw vy vs t v v v9 A 9 B 9 C 0.5 t 0.5 t 0.5 t -6 -6 -6 v v 9 9 E D 0.5 t 0.5 t -6 -6 Physics 101: Lecture 3, Pg 11
- 12. Dropped Ball xA ball is dropped for a height of twometers above the ground. t v Draw v vs t Draw x vs t t Draw a vs t a t Physics 101: Lecture 3, Pg 12
- 13. Tossed Ball•A ball is tossed from the ground up a height of twometers above the ground. And falls back down y x Draw v vs t v v v 9 A 9 B 9 C 1 t 1 t 1 t -6 -6 -6 v v 9 9 D E 1 t 1 t -6 -6 Physics 101: Lecture 3, Pg 13
- 14. Tossed Ball x•A ball is tossed from the ground up aheight of two meters above the ground. tAnd falls back down v Draw v vs t Draw x vs t t Draw a vs t a t Physics 101: Lecture 3, Pg 14
- 15. ACTA ball is thrown straight up in the air and returns to itsinitial 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. Vave = DY/Dt = (Yf – Yi) / (tf – ti) = 0 aave = DV/Dt = (Vf – Vi) / (tf – ti) Not 0 since Vf and Vi are not the same ! 101: Lecture 3, Pg 15 Physics
- 16. Relative Velocity (first pass)You are on a train traveling 40 mph North. If you walk 5 mph toward the front of the train, what is your speed relative to the ground?A) 45 mph B) 40 mph C) 35 mph 40 mph N + 5 mph N = 45 mph N 40 5 45 Physics 101: Lecture 3, Pg 16
- 17. Relative VelocityYou are on a train traveling 40 mph North. If you walk 5 mph toward the rear of the train, what is your speed relative to the ground?A) 45 mph B) 40 mph C) 35 mph 40 mph N - 5 mph N = 35 mph N 40 5 35 Physics 101: Lecture 3, Pg 17
- 18. Relative VelocityYou are on a train traveling 40 mph North. If you walk 5 mph sideways across the car, what is your speed relative to the ground?A) < 40 mph B) 40 mph C) >40 mph 40 mph N + 5 mph W = 41 mph N 5 | v | 40 2 5 2 40 Physics 101: Lecture 3, Pg 18
- 19. Relative Velocity Sometimes your velocity is known relative to a reference frame that is moving relative to the earth. Example 1: A person moving relative to a train, which is moving relative to the ground. Example 2: a plane moving relative to air, which is then moving relative to the ground. These velocities are related by vector addition: » vac is the velocity of the object relative to the groundv ac v ab v bc » vab is the velocity of the object relative to a moving reference frame » vbc is the velocity of the moving reference frame relative to the ground Physics 101: Lecture 3, Pg 19
- 20. Tractor Demo 1Which direction should I point the tractor to get it across the table fastest?A) 30 degrees leftB) Straight acrossC) 30 degrees right 1 2 3 Physics 101: Lecture 3, Pg 20
- 21. Tractor Demo (moving table) Which direction should I point the tractor to get it across the table fastest?A) 30 degrees leftB) Straight acrossC) 30 degrees right 1 2 3 Physics 101: Lecture 3, Pg 21
- 22. Summary of Concepts kinematics: A description of motion position: your coordinates displacement: Dx = change of position velocity: rate of change of position average : Dx/Dt instantaneous: slope of x vs. t acceleration: rate of change of velocity average: Dv/Dt instantaneous: slope of v vs. t relative velocity: vac = vab + vbc Physics 101: Lecture 3, Pg 22

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment