Boris AustinGetty Images Constant Velocity Constant Acceleration Free
Boris Austin/Getty Images • • Constant Velocity Constant Acceleration Free Fall Graphical Analysis of Motion College Physics for the AP® Physics 1 Presentation Created by Martin Kirby
Finding the slope of a linear function. Here is a quick reminder about finding the slope of a linear function. 6 y e p o sl 4 2 ∆x 0 2 1 ∆: delta For example: value ∆y 2 3 x 4 ∆ means ‘the change in…’ ∆y = y final value– y initial COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Finding the slope at a point on a non-linear function Draw a line that has the same slope as your chosen point on the curved graph. 6 4 2 0 2 y n o t n ion i o p nct n e s u f o r h C inea l n no 1 2 e p o sl ∆y ∆x 3 x 4 The slope of the blue line (a tangent line) is the same as the slope of the non-linear function at the chosen point. COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Linear Motion If an object is moving in a straight line we say that it is in linear motion. COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Non-Linear Motion If an object is not moving in a straight line we say that it is in non-linear motion. COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
∆x: Displacement The displacement, ∆x, of an object is the difference between the object’s final and initial position. x 1 ∆x x 1 x 2 ∆x x 2 The difference between the starting and ending positions of the two objects above is the same. Their displacements, ∆x, are the same. ∆x = x 2 – x 1 COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
d: Distance The distance, d, that an object moves is the length of the path it has traveled. The direction in which it is traveling is irrelevant. xo ∆x d xo x ∆x d x The distance, d, traveled by the two objects above are very different. However, their displacements, ∆x, are the same. COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Comparison of ∆x and d Displacement, ∆x • has magnitude and direction, • what happens between the start and end positions is irrelevant, and • can be negative or positive Distance, d • has magnitude only, • what happens between the start and end positions is relevant, • has no sign (+ or -), • cannot decrease COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Which is more useful? Which statement is more useful? Your prize of 2 kg of gold is at the end of a 3. 2 km walk from here. d, distance (only magnitude) Your prize of 2 kg of gold is exactly 3. 2 km east from here. ∆x, displacement (magnitude and direction) COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Average Speed and Average Velocity vaverage : average speed vaverage = distance change in time d = ∆t unit: m/s vaverage, x : average velocity displacement change in position ∆x vaverage, x = change in time = ∆t unit: m/s COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Average Velocity, vaverage, x ∆x = +6 m 0 m -2 m x (m) 8 6 4 2 0 -2 +4 m +2 m x +6 m ∆x = +6 m 1 2 3 t (s) ∆t = 3 s COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Average Speed, vaverage d=6 m 0 m -2 m x (m) 8 6 4 2 0 -2 4 m 2 m x 6 m d=6 m 1 2 3 t (s) ∆t = 3 s COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Average Velocity, vaverage, x 2 ∆x = +4 m x (m) 8 6 4 2 0 -2 +4 m +2 m 0 m -2 m x +6 m ∆x = +4 m 1 2 3 4 t (s) ∆t = 4 s COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Average Speed, vaverage d 2 = 2 m d 1 = 6 m – 2 m 10 8 6 4 2 0 +4 m +2 m 0 m d (m) 2 x +6 m d 2 = 2 m d 1 = 6 m t (s) 1 2 3 4 ∆t = 4 s COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Average Velocity, vaverage, x 3 ∆x = – 20 m 0 m +60 m +40 m +20 m x +80 m ∆t (s) 80 x (m) 60 40 20 0 -20 1 2 3 4 5 6 7 ∆t = 10 s 8 9 3. 00 1. 00 2. 00 t (s) 2. 00 10 ∆x = -20 m 1. 00 xfinal- xinitial 10. 00 s ∆x (m) +40 -20 +60 – 20 0 – 80 – 20 m COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Average Speed, vaverage – 20 m d (m) 0 m 240 200 160 120 80 40 0 +60 m +40 m +20 m 3 x +80 m ∆t (s) 3. 00 40 1. 00 20 60 1. 00 ∆x = 220 m 2. 00 20 2. 00 0 80 1. 00 Total 10. 00 s 220 m ∆t = 10 s 1 2 3 4 t (s) 5 6 d (m) 7 8 9 10 COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
The slope of position-time graph is velocity. x ∆t ∆x rise ∆x slope = run = ∆t = velocity t The slope of a position-time graph = velocity Zero velocity x positive velocity negative velocity t COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Instantaneous Velocity and Average Velocity 1 Instantaneous velocity: Slope at this instant is positive The slope of this x-t graph is constantly changing. Therefore, this graph represents a velocity that’s constantly changing. To find the velocity at an instant (instantaneous velocity), find the slope of the graph at that instant. x Slope at this instant is negative Slope at this instant is zero Instantaneous velocity, v is the velocity at an instant. Average velocity, vaverage, x is the average velocity between two instants. t COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Instantaneous Velocity and Average Velocity 2 The slope between these two instants is ∆x / ∆t = – 2 m / 4 s = – 0. 5 m/s The slope at this instant is ∆x / ∆t = +4 m / 8 s = + 0. 5 m/s. At this one instant, The instantaneous velocity is + 0. 5 m/s Between these two instants the average velocity is – 0. 5 m/s. 8 8 ∆x = +4 m 6 6 ∆x = -2 m 4 ∆t = 8 s 4 ∆t = 4 s 2 2 x (m) 0 2 4 6 8 t (s) 10 12 14 16 COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Velocity-Time Graphs 6 – 2 m 0 m 2 0 – 2 4 m 2 m 6 m 8 m ∆x = +8 m 4 X (m) – 4 m 1 Slope = +8 m = + 2 m/s 4 s 1 2 3 t (s) 4 4 v (m/s) Velocity = + 2 m/s 3 2 1 0 1 2 3 t (s) COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE 4
Velocity-Time Graphs x=0 x constant t zero slope x constant t zero velocity x constant t negative slope v v constant t positive velocity x=0 constant t negative slope constant t positive slope v v x=0 x 2 constant t negative velocity COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Velocity-Time Graphs 0 m 16 X 2 m 4 m 8 m 6 m 10 m 3 12 m 14 m 16 m s / m 4 12 8 (m) 4 0 = +1 m/s Slope = 0 m/s Slope 1 /s m 2 + pe = Slo 2 4 = e p Slo /s m +3 Slo + = e p 3 t (s) 4 3 2 (m/s) 1 v 0 1 2 COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Acceleration average acceleration aaverage = change in velocity change in time ∆v = ∆t unit: m/s 2 COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Graphs: x-t and v-t and a-t 6 4 2 0 (m) – 2 X v 4 3 2 1 0 a 3 2 1 0 (m/s) 4 (m/s 2) – 4 m – 2 m 0 m 4 m 2 m 1 6 m 8 m 1 2 3 t (s) 4 COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Graphs: x-t and v-t and a-t 0 m 16 12 8 4 0 (m) 8 6 4 2 (m/s) 0 x 2 m 4 m 6 m 8 m 10 m 12 m a (m/s 2) 14 m 16 m 1 2 3 t (s) 4 v 4 3 2 1 0 2 COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Graphs: v-t and a-t 0 m 8 v 2 m 4 m 6 m 8 m 10 m 3 12 m 14 m 16 m 6 4 (m/s) 2 0 4 3 2 (m/s 2) 1 a 0 1 aavg= ∆ velocity ∆ time 1 2 3 t (s) 4 ∆v +8 m/s = +2 m/s 2 = = 4 s ∆t 2 3 COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Speeding Up and Slowing Down 4 2 0 (m/s) -2 -4 v 4 2 0 -2 2 (m/s ) -4 a 2 2 slowing down…. accelerationisisstill – 2 m/s speeding up…. acceleration – 2 m/s The terms speeding up and slowing down give only magnitude 3 4 2 1 (s) information. t Acceleration gives magnitude and 3 2 1 (s) 4 direction information. COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE t
Note well! • Note: if velocity and acceleration are in the same direction the object is speeding up. • Note: if velocity and acceleration are in opposite directions then you are slowing down. • Note: If an object’s instantaneous velocity is zero, and the object is accelerating, then the object is changing direction at that instant! • Note: Don’t describe a changing velocity as deceleration…always describe it as either negative acceleration or positive acceleration. COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Area under a v-t Graph = ∆x 1 v +3 m/s Area = (+3 m/s)(4 s) = +12 m Area = ∆x v +3 m/s +1 m/s t 4 s Area = (+1 m/s)(4 s) + ½(4 s)(+2 m/s) = +4 m + 4 m = +8 m Area = ∆x 4 s t ∆x Vavg, x = ∆t ∆x = vavg, x ∆t ∆x = (+3 m/s)(4 s) ∆x = +12 m ∆x = vo∆t + ½a∆t 2 ∆x = vo∆t + ½(∆v/∆t)∆t 2 ∆x = (+1)(4) + ½(+2/4)(4)2 ∆x = +4 m + 4 m ∆x = +8 m COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Area under a v-t Graph = ∆x v (m/s) 2 4 2 0 A -2 1 B 2 C 3 D 4 t (s) -4 ∆x = total area under v-t graph ∆x = area A + area B + area C + area D ∆x = ½(+4 m/s)(1 s) + ½(1 s)(– 4 m/s) + (1 s)(– 4 m/s) + ½(1 s)(– 4 m/s) ∆x = (+2 m) + (– 4 m) + (– 2 m) ∆x = – 6 m Note that area A and area B ‘cancel each other out’. COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Kinematics Equations average velocity: acceleration: Not provided on AP equation sheet. xo : initial position x : final position vo : initial velocity v : final velocity to : initial velocity t : final velocity a : acceleration As shown on AP equation sheet. ∆x = x – xo ∆v = v – vo ∆t = t – to COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Kinematics Problem A cloud has a constant acceleration of +4 m/s 2 to the West, and is initially moving East at 8 m/s. After 3 s have passed, find the cloud’s displacement vo = +8 m/s v = ∆x = ? a = 4 m/s 2 ∆t = 3 s x = xo + vxot + ½ax∆t 2 x = 0 + (-8)(3) + ½(+4)(3)2 x = -24 m + 18 m x=-6 m x = 6 m to the east COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
g: Magnitude of the Acceleration due to Gravity The magnitude of the acceleration due to gravity, near Earth ≈ 9. 8 m/s 2 +y -y -y If the positive y direction points upward from Earth: The acceleration due to gravity is: a = –g ≈ – 9. 8 m/s 2. a = – g = – 9. 8 m/s 2 If the positive y direction points downward to Earth : The acceleration due to gravity is: a = +g ≈ +9. 8 m/s 2. y a = + g = + 9. 8 m/s 2 +y COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
Chapter 2 Summary Linear motion (1 -dimensional motion) is motion in a straight line. ∆x = x – xo : difference between position x at two different times ∆v = v – vo : difference between velocity v at two different times Slope of an x–t graph equals velocity. Slope of a v–t graph equals acceleration. Vaverage, x = ∆x /∆t aaverage = ∆v /∆t x = xo + vo∆t + ½a∆t 2 v 2 = vo 2 + 2 a∆x An object is in free fall if its motion is influenced by gravity alone. COLLEGE PHYSICS FOR THE AP® PHYSICS 1 COURSE
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