Lecture 2- Linear Motion Chapter 10
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1 1 / 37 Lecture 2- Linear Motion Chapter 10 Instructor: Prof. Noronha-Hostler Course Administrator: Prof. Roy Montalvo PHY-123 ANALYTICAL PHYSICS IA Phys- 123 Sep. 12 th, 2018
2 Contact Already read the syllabus first! Your TA is your first point of contact (one exception, office of disability letters) If your TA can t answer your question, they will contact the correct person Josh Gier, the Northeast District Manager, has offered his address for student feedback on WileyPLUS and pricing: jgier@wiley.com 2 / 37
3 3 / 37 Exam schedule posted Look Underneath the section "Announcements"
4 4 / 37 Objectives Difference between speed vs. velocity Average speed vs. instantaneous speed Understand acceleration Predict position and/or velocity when acceleration is constant Read and understand graphs depicting movement
5 5 / 37 What is linear motion? Linear motion Car driving down a flat stretch of highway Ball being thrown into the air Sled down a hill Rocket launch Person walking down the street Non-linear motion Clock Ball rolling down a hill Satellite orbiting the Earth
6 6 / 37 Position Origin x(m)
7 7 / 37 Positive displacement + displacement x(m)
8 8 / 37 Negative displacement - displacement x(m)
9 9 / 37 Displacement Start at A, walk to B: x = x B x A = 3 ( 2) = 5 m Start at B, walk to A: x = x A x B = ( 2) 3 = 5 m A Δx =5 B x(m)
10 Displacement as a vector Displacement, x, always has an associated direction (the corresponding sign) and magnitude (absolute value x ). Vector Quantity that has both a direction and magnitude. magnitude is the distance from the original position to final position direction from original position to final position e.g. plus or minus sign for linear motion 10 / 37
11 11 / 37 Stationary
12 12 / 37 Motion Dimensional analysis [length/time] A B
13 Average velocity A B v avg = x t = x B x A t B t A = 20m 5s = 4m/s 13 / 37
14 14 / 37 Average velocity (vector) 3.0s B 8.0s A v avg = x t = x A x B t A t B = 20m 5s = 4m/s
15 15 / 37 Average velocity- slope Slope of the straight line between initial and final position= average velocity
16 Average speed (scalar) A B s = total distance t = x B x A t B t A = 20m 5s = 4m/s 16 / 37
17 Average speed (scalar) 3.0s B 8.0s A s = total distance t = x A x B t A t B = 20m 5s = 4m/s 17 / 37
18 A particle travels along a curved path between two points A and B as shown. Complete the following statement: The displacement of the particle does not depend on a) the location of A. b) the location of B. c) the direction of A from B. d) the distance traveled from A to B. e) the shortest distance between A and B.
19 18 / 37 Instantaneous Velocity x v = lim t 0 t = dx dt Velocity obtained as t is shrunk closer to zero. Vector quantity
20 19 / 37 Instantaneous speed v Magnitude of the instantaneous velocity
21 A motorcycle travels due south covering a total distance of 80.0 kilometers in 60.0 minutes. Which one of the following statements concerning this situation is necessarily true? a) The velocity of the motorcycle is constant. b) The acceleration of the motorcycle must be non-zero. c) The motorcycle traveled 40.0 kilometers during the first 30.0 minutes. d) The speed of the motorcycle must be 80.0 kilometers per hour throughout the entire trip. e) The average velocity of the motorcycle is 80.0 kilometers per hour, due south.
22 The speedometer on a car s dashboard measures which of the following quantities? a) average speed b) average velocity c) average acceleration d) instantaneous velocity e) instantaneous acceleration
23 20 / 37 Acceleration-Change in velocity Average Acceleration a avg = v 2 v 1 t 2 t 1 = v t Instantaneous acceleration a = dv dt Dimensional Analysis [ length/time 2]
24 21 / 37 Acceleration vs position recall v = dx dt a = dv dt = d dt ( ) dx dt (1) (2) = d 2 x dt 2 (3)
25 22 / 37 Slopes (demo)
26 23 / 37 g Units (gravity) Magnitude of acceleration of a falling object near Earth s surface 1g = 9.8m/s 2 Example roller coaster 3g.
27 24 / 37 Example Given the position x = t 12t 2 What is the particle s velocity* v(t) and acceleration* a(t)? Velocity v = dx (4) dt = d ( t 12t 2) (5) dt = t (6) = t (7) *unless specified always assume instantaneous velocity or acceleration
28 25 / 37 Example Given the position x = t 12t 2 What is the particle s velocity* v(t) and acceleration* a(t)? Acceleration a = dv dt (8) = d (0.5 24t) dt (9) = 24 (10) Acceleration is constant in this problem! *unless specified always assume instantaneous velocity or acceleration
29 26 / 37 Example Given the position x = t 12t 2 When is v = 0? v(t) = 0 = t (11) 24t = 0.5 (12) t = 0.021s (13) Notice the significant figures!
30 27 / 37 Special case: Velocity with a=const Because a=constant, the average acceleration is equal to a(t) a = a avg = v v 0 (14) t 0 v = v 0 + at (15)
31 28 / 37 Special case: Position with a=const v avg = x x 0 (16) t 0 x x 0 = v avg t (17) For a=const, the average velocity is the just the average between the initial and final velocity v avg = 0.5 (v 0 + v) Recall, v = v 0 + at (last slide), then Substituting Eq. (19) into Eq. (17) v avg = 0.5 (v 0 + v 0 + at) (18) = v at (19) x x 0 = v 0 t + 0.5at 2
32 29 / 37 Equations of motion for constant acceleration Choose the equation where the only unknown variable is the variable requested in the problem Equation Missing variables v = v 0 + at x x 0 x x 0 = v 0 t + 0.5at 2 v v 2 = v a(x x 0) t x x 0 = 0.5 (v 0 + v) t a x x 0 = vt 0.5at 2 v 0 See book for derivations
33 30 / 37 Rolling ball (demo) Regardless of the mass, a ball rolling down a slope should follow this equation (to a good approximation) x x 0 = v 0 t + 0.5at 2 Assuming v 0 = 0 and x 0 = 0 (set at the origin), x = 0.5at 2
34 31 / 37 Example: Braking at a light Can the car stop in time? A BMW car traveling 90. km/hour brakes at a = 9.8m/s 2 for a red light. The driver begins braking 30. m before the light, will the car stop in time? Which equation should we use? v = v 0 + at (20) x x 0 = v 0 t + 0.5at 2 (21) v 2 = v a(x x 0) (22) x x 0 = 0.5 (v 0 + v) t (23) x x 0 = vt 0.5at 2 (24)
35 32 / 37 Example: Braking at a light Can the car stop in time? A BMW car traveling 90. km/hour brakes at a = 9.8m/s 2 for a red light. The driver begins braking 30. m before the light, will the car stop in time? We know: a = 9.8m/s 2 and v 0 = 90m/hour. We don t know the time. We re looking for what x we need for v = 0.
36 33 / 37 Example: Braking at a light Can the car stop in time? A BMW car traveling 90. km/hour brakes at a = 9.8m/s 2 for a red light. The driver begins braking 30. m before the light, will the car stop in time? We know: a = 9.8m/s 2 and v 0 = 90.m/hour. We don t know the time. We re looking for what x we need for v = 0. v 2 = v a(x x 0)
37 Example: Braking at a light Can the car stop in time? A BMW car traveling 90. km/hour brakes at a = 9.8m/s 2 for a red light. The driver begins braking 30. m before the light, will the car stop in time? First, we convert 90. km hour Then, km 1 hour = minute hour 60. minutes 60.seconds 1000m 1 km = 25m/s 0 = (25m/s) ( 9.8m/s 2 )x (25) x = 32m (26) Since 32m > 30m, No the car does not stop in time! 34 / 37
38 35 / 37 Free Fall acceleration (demo) v = v 0 + at (27) = (m/s 2 )t (28) = 9.5(m/s 2 )t (29)
39 36 / 37 Calculus version of x(t), v(t), and a=const
40 Which one of the following position versus time graphs depicts an object moving with a negative constant velocity?
41 Which one of the following quantities can be determined from the slope of a position versus time graph for an object in motion? a) position b) velocity c) acceleration d) distance traveled e) displacement
42 Complete the following statement: For an object moving at constant velocity, the distance traveled a) increases for each second that the object moves. b) is the same regardless of the time that the object moves. c) is the same for each second that the object moves. d) cannot be determined, even if the elapsed time is known. e) decreases for each second that the object moves.
43 37 / 37 Next Week Vectors
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