Chapter 2 Motion Along A Straight Line Kinematics: Description of Motion Motion in one dimension (1-D) Motion of point particles Treat larger objects as particles center of mass Chapter 2 Motion in 1-D 1
Position - x x Origin: Reference point Position - x: distance to the right of origin What if the particle is to the left of the origin? x is negative x <0 Position is a vector It has a magnitude and direction The length, x, is the magnitude The sign is the direction Chapter 2 Motion in 1-D 2
Examples of Position x = -25 m x = +7.5 m x = +30 m -30 m -20 m -10 m 0 10 m 20 m 30 m origin Chapter 2 Motion in 1-D 3
Displacement - x Change in position ( represents change in ) x x = x final f - x- i x initial Displacement, x, is a vector Displacement to the right is positive Displacement to the left is negative x > 0 x < 0 The length is the magnitude: x Chapter 2 Motion in 1-D 4
Examples of Displacement x = x f - x i 30.0m - 7.5m = 22.5 m x i = +7.5 m x f = +30.0 m -30 m -20 m -10 m 0 10 m 20 m 30 m Chapter 2 Motion in 1-D 5
Examples of Displacement x = x f - x i 5m - 25m = -20 m x f = +5 m x i = +25 m -30 m -20 m -10 m 0 10 m 20 m 30 m Chapter 2 Motion in 1-D 6
Examples of Displacement x = x f - x i -10m - (-35m) = +25 m x i = -35 m x f = -10 m -30 m -20 m -10 m 0 10 m 20 m 30 m Chapter 2 Motion in 1-D 7
The difference between displacement and total distance Consider a round trip 10 miles there 10 miles back again The total distance traveled is 20 miles The displacement, x = x f - x i In a round trip, the final position is the same as the starting position x f = x i So the displacement, x = 0 The displacement and the total distance traveled are not necessarily the same! Chapter 2 Motion in 1-D 8
Another Example x = -12 km 20 km to the right Then 32 km to the left The total distance traveled is 20 km + 32 km = 52 km The displacement is, x = -12 km The total distance traveled is always x Chapter 2 Motion in 1-D 9
Velocity - v Average Velocity, v avg v avg = x/ t = displacement/time interval time interval, t = t f - t i, how much time it takes to undergo the displacement Example: v avg = x/ t x = + 140 m t = 4.0 s = +140m/4.0s = +35 m/s SI unit of velocity is m/s Chapter 2 Motion in 1-D 10
Example: x = -130 miles t = 2.0 hours v avg = x/ t = -130miles/2.0hours = -65miles/hour Velocity is a vector With magnitude (+65 miles/hour) and direction (left or -) Example: A round trip 10 miles there 10 miles back again On a round trip x = 0. So v avg = x/ t also = 0 Chapter 2 Motion in 1-D 11
The position, x is a function of time x =x(t) This means that at every moment of time there is a unique position for every object At, t = -2s x = -40 m At, t = 0s x = -25 m At, t = 2s x = -5 m At, t = 4s x = +20 m -40 m x(-2s) = -40m -30 m -20 m -10 m 0 10 m 20 m 30 m x(0s) = -25m x(2s) = -5m x(4s) = +20m And, of course, the car has a position at every moment in between Chapter 2 Motion in 1-D 12
We can represent the function x(t) with a graph 30 m Position (m) 20 m 10 m x(t) x(+2.0s) = +15m 0-4.0 s -3.0 s -2.0 s -1.0 s 0 1.0 s 2.0 s 3.0 s 4.0 s -10 m Time (s) x(0) = -12 m x(-2.0s) = -16 m -20 m Chapter 2 Motion in 1-D 13-30 m
500 400 Average Velocity and Slope x f = 400m x(t) Position (m) 300 200 v avg = x/ t 300m/40s 7.5m/s v avg = slope x = x300 400m-100m f -x i m 100 x i = 100m t = t50s 40s f -t i - 10s 0 t i = 10s t f = 50s 10 0 10 20 30 40 50 60 70 Time (s) -100 Chapter 2 Motion in 1-D 14
500 400 Letting the time interval shrink t = 40s x(t) Position (m) 300 200 100 0 t 10 0 10 20 30 40 50 60 70 Time (s) -100 Chapter 2 Motion in 1-D 15
500 400 Letting the time interval shrink t = 30s x(t) Position (m) 300 200 100 0 t 10 0 10 20 30 40 50 60 70 Time (s) -100 Chapter 2 Motion in 1-D 16
500 400 Letting the time interval shrink t = 20s x(t) Position (m) 300 200 100 0 t 10 0 10 20 30 40 50 60 70 Time (s) -100 Chapter 2 Motion in 1-D 17
500 400 Letting the time interval shrink t = 10s x(t) Position (m) 300 200 100 0 t 10 0 10 20 30 40 50 60 70 Time (s) -100 Chapter 2 Motion in 1-D 18
500 400 Letting the time interval shrink t = 5s x(t) Position (m) 300 200 100 0 t 10 0 10 20 30 40 50 60 70 Time (s) -100 Chapter 2 Motion in 1-D 19
Position (m) 500 400 300 200 Letting the time interval shrink to zero t 0 lim t 0 x/ t = dx/dt v avg x(t) Tangent Line The slope of the Tangent line = dx/dt 100 0 10 0 10 20 30 40 50 60 70 Time (s) -100 Chapter 2 Motion in 1-D 20
Instantaneous Velocity - v v = dx/dt The slope of the tangent line to the graph of x(t) Example x = (5m/s)t (dt/dt = 1) v = dx/dt = d[(5m/s)t]/dt = (5m/s) dt/dt = 5m/s 1 = 5m/s Chapter 2 Motion in 1-D 21
50 x = (5m/s)t Position (m) 40 30 Slope = 5m/s = dx/dt = v Constant Velocity 20 10 0-10 0 1 2 3 4 5 6 7 Time (s) Chapter 2 Motion in 1-D 22
Homework 1 Extention Due Wednesday, 10/9 Chapter 2 Motion in 1-D 23
Recap Position - x Displacement - x = x f -x i Average Velocity - v avg = x/ t Instantaneous Velocity - v = dx/dt Chapter 2 Motion in 1-D 24
Position (m) 50 40 30 20 10 0 Example: x= (1m/s 2 )t 2 dx/dt = d[ (1m/s 2 )t 2 ]/dt = (1m/s 2 ) d[ t 2 ]/dt d[ t 2 ]/dt = 2t = (1m/s 2 ) 2t v(t) = (2m/s 2 ) t Not Constant Slope increases with time t x 0 0 1s 1m 2s 4m 3s 9m 4s 16m 5s 25m 6s 36m 7s 49m -10 0 1 2 3 4 5 6 7 Time (s) Chapter 2 Motion in 1-D Wed. 9/29 25
20 Velocity (m/s) 16 12 8 4 0 v(t) = (2m/s 2 )t Slope = 2m/s 2 t v 0 0 1s 2m/s 2s 4m/s 3s 6m/s 4s 8m/s 5s 10m/s 6s 12m/s 7s 14m/s -4 0 1 2 3 4 5 6 7 Time (s) Chapter 2 Motion in 1-D 26
Speed - v The instantaneous velocity is a vector with magnitude and direction The instantaneous speed is the magnitude of the instantaneous velocity Speed = v For instance: v = + 50 miles/hour Speed = v = +50 miles/hour v = -50 miles/hour Speed = v = +50 miles/hour Chapter 2 Motion in 1-D 27
Position (m) 500 400 300 200 100 0-100 Negative velocity motion in -x direction Direction of velocity The direction of the velocity is the direction of motion Zero velocity Turning point Positive velocity motion in +x direction 0 10 20 30 40 50 60 70 Time (s) Chapter 2 Motion in 1-D 28
Average Speed - v avg The average speed, v avg = total distance traveled/time interval The average speed is not the magnitude of the average velocity For instance a round trip: 10 miles there t = 40 min 10 miles back again V avg = x/ t = 0/40min = 0 v avg = total distance traveled/ t = 20miles/40min = 0.50 miles/min = 30 miles/hour Chapter 2 Motion in 1-D 29
Acceleration - a Acceleration is the rate of change of the velocity Average acceleration, a avg = v/ t = (v f -v i )/(t f -t i ) = change in velocity/time interval Example: v i = 0 t = 100 s v f = +20m/s a avg = v/ t = (v f -v i )/(t f -t i ) = (+20 m/s -0)/100s = (20 m/s)/100s = 0.20 (m/s)/s = 0.20 m/s 2 On average the velocity increases by 0.20 m/s every second Chapter 2 Motion in 1-D 30
Acceleration is a vector Acceleration has magnitude and direction Example: v i = +20 m/s t = 1.0 min v f = -10m/s a avg = v/ t = (v f -v i )/(t f -t i ) = (-10 m/s -(+20))/1.0 min = (-30 m/s)/60s = -0.50 (m/s)/s = -0.50 m/s 2 The acceleration is in the -x direction The magnitude of the acceleration, a = + 0.50 m/s 2 The velocity is becoming more negative (left) by 0.50 m/s 2 every second Chapter 2 Motion in 1-D 31
Acceleration / Deceleration Deceleration is decreasing speed Deceleration is not the same as negative acceleration Deceleration occurs when the direction of the instantaneous acceleration is opposite the direction of the instantaneous velocity Acceleration - Velocity + or Velocity - Acceleration + When the acceleration and the velocity are in the same direction the speed increases Acceleration + Velocity + or Velocity - Acceleration - Chapter 2 Motion in 1-D 32
Constant Acceleration x = x o + v o t + (1/2)at 2 - Position as a function of time x o, v o, and a are constants x o is the initial position, v o is the initial velocity, a is the acceleration x o = x(0) v o = v(0) a = constant Velocity v = dx/dt = d[x o + v o t + (1/2)at 2 ]/dt = d[x o ]/dt + d[ v o t ]/dt + d[(1/2)at 2 ]/dt = 0 + v o d[t ]/dt + (1/2)ad[t 2 ]/dt = 0 + v o 1 + (1/2)a (2t) = 0 + v o + at v = v o + at - Velocity as function of time Chapter 2 Motion in 1-D 33
Constant Acceleration v = v o + at - Velocity as function of time Acceleration a = dv/dt = d[v o + at]/dt = d[v o ]/dt + d[ at ]/dt = 0 + a d[t ]/dt = 0 + a 1 a = a = constant Chapter 2 Motion in 1-D 34
Constant Acceleration x = x o + v o t + (1/2)at 2 at t = 0 x(t) = x o + v o t + (1/2)at 2 v = v o + at a = a = constant v(t) = v o + at x(0) = x o + v o (0) + (1/2)a(0) 2 x(0) = x o + 0 + 0 x(0) = x o v(0) = v o + a(0) v(0) = v o + 0 v(0) = v o a(t) = a a(0) = a Chapter 2 Motion in 1-D 35
500 Constant Acceleration x(t) = x o + v o t + (1/2)at 2 Position (m) 400 300 x o = 325m 200 100 v = 0 Constant Acceleration: x(t) is a parabola 0-100 0 10 20 30 40 50 60 70 Time (s) Chapter 2 Motion in 1-D 36
60 Constant Acceleration v(t) = v o + at Velocity (m/s) 40 20 Constant Acceleration: v(t) is a straight line Slope = acceleration = constant = (30m/s)/28s = 1.1m/s 2 0-20 0 10 20 30 40 50 60 70 Time (s) -40 v o = -30 m/s Chapter 2 Motion in 1-D 37
2.0 Constant Acceleration a(t) = a=constant 1.6 Constant Acceleration: a(t) is a flat line 1.2 Acceleration (m/s 2 ) 0.8 0.4 a = +1.1 m/s 2 x(t) = 325m x o + - (-30m/s)t (30m/s)t v o t + (0.55m/s (1/2)(1.1m/s (1/2)at 2 )t 22 2 )t 2 v(t) = -30m/s v o + (1.1m/s at 2 )t 0-0.4 0 10 20 30 40 50 60 70 Time (s) Mon. 9/30 Chapter 2 Motion in 1-D 38
Negative Constant Acceleration x = x o + v o t + (1/2)at 2 v = v o + at a = a = constant Ex: xo = 10 m, vo = 5 m/s, a = -0.50m/s^2 v vs. t slopes down v=0 at 10s Inverted Parabola xmax = 35m Chapter 2 Motion in 1-D 39
Free fall - Gravity Free fall is accelerated motion Free fall acceleration is directed toward the center of the Earth All bodies fall with the same acceleration (when air friction is negligible) Free fall near the surface of the Earth is motion with constant acceleration The magnitude of the acceleration due to gravity near the surface of the Earth, g = 9.8 m/s 2 Chapter 2 Motion in 1-D 40
Free fall - Gravity 14 Vertical Position vs. Time y = 12.962m - (0.8868m/s) t -(4.9152m/s^2) t 2 12 10 y (m) 8 6 4 2 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 time (s) Chapter 2 Motion in 1-D 41
Free fall - Gravity y x = yx o + v o t + (1/2)at 2 y is the vertical position From the previous graph: y = 12.962m - (0.8868m/s) t -(4.9152m/ m/s 2 ) t 2 y (m) 160 120 80 y o = 12.962m v o = -0.8868 m/s (1/2)a = -4.9152m/s 2 a = 2 x ( -4.9152m/s 2 ) = -9.8304m/s 2 2 a = 9.8 m/s 2 = g Direction of a is down 40 0-40 Chapter 2 Motion in 1-D 42
Free fall is an example of motion with constant acceleration Constant Acceleration Free Fall x = x o + v o t + (1/2)at 2 v = v o + at a = a = constant x = v i t + (1/2)a( t) 2 v = a t v avg = (v i +v f )/2 = x / t v 2 f = v i2 +2a x y = y o + v o t - (1/2)gt 2 v = v o - gt a = -g = -9.8 m/s 2 y = v i t - (1/2)g( t) 2 v = -g t v avg = (v i +v f )/2 = y / t v 2 f = v 2 i - 2g y Chapter 2 Motion in 1-D 43