# KINEMATICS in One Dimension

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KINEMATICS in One Dimension
REPRESENTING MOTION KINEMATICS in One Dimension “To understand motion is to understand nature.” Leonardo da Vinci

Study of motion, force and energy
MECHANICS Study of motion, force and energy Kinematics How objects move Dynamics Why objects move

Kinematics Objectives
Represent motion through the use of words, motion diagrams, graphs, and mathematical models. Use the terms position, distance, displacement, and time interval in a scientific manner to describe motion. 3

Motion Motion is instinctive
Eyes will notice moving objects more readily than stationary ones Object changes position Motion can occur in many directions and paths

Representing Motion A description of motion relates PLACE and TIME.
Answers the questions WHERE? and WHEN? PLACE TIME

Motion Diagram & Particle Model
Simplified version of a motion diagram in which the object in motion is replaced by a series of single points Size of object must be much less than the distance it moves

Describe motion of the car…
Draw a particle model…

How are the two particle models different? Describe the motion of each.

Reference Frames Any measurement of position, distance or speed must be made with respect to a frame of reference 80 km/h

Coordinate System Tells you the location of the zero point of the variable you are studying and the direction in which the values of the variable increase. ORIGIN The point at which both variables have the value zero

Distance and Displacement
Distance, d – total ground covered Displacement, Dx – change in position of an object (position is measured from the origin of a chosen coordinate system)

Distance and Displacement
Example - A car travels 400 km from Livingston to Philadelphia and then back 200 km to Trenton. What is the displacement of the car? What distance did the car travel? X = 100 200 300 400 km Livingston Trenton Philadelphia displacement distance

Distance and Displacement
Example - A car travels 400 km from Livingston to Philadelphia and then back 200 km to Trenton. What is the displacement of the car? What distance did the car travel? X = 100 200 300 400 km Livingston Trenton Philadelphia displacement distance

Displacement Has magnitude (size) and direction. It is a VECTOR
Vectors are represented by arrows x(m) 1 2 3 -1

Has magnitude and direction. It is a VECTOR
Displacement Has magnitude and direction. It is a VECTOR x(m) 1 2 3 -1 15

Average Speed and Average Velocity
Physics C 1-D Motion Average Speed and Average Velocity Average speed describes how fast a particle is moving. It is calculated by: Average velocity describes how fast the displacement is changing with respect to time: always positive Scalar (has magnitude only) sign gives direction in 1 Dimension Vector (has magnitude and direction) Average speed and average velocity often have the same magnitude, but not always Bertrand

Example - A car travels 400 km from Philadelphia to Livingston in 2 hours and then back 200 km to Trenton in 1 hr. What is the car’s speed and velocity? Dx X = 100 200 300 400 km Livingston Trenton Philadelphia Average speed Average velocity

Example The position-time graph shows the progress of two runners, A and B. When does runner B pass runner A? Where does runner B pass runner A? What is the starting position for runner A? runner B? After 10 hrs, what is the average velocity of runner A? After 10 hrs, what is the average velocity of runner B? If the finish line is at 40 km, who won the race? 18

Analyzing Graphs UNITS are the key to analyzing graphs
When analyzing graphs always check for the following two things: Slope: Look at the units of the slope to see if it corresponds to a physically meaningful measurement. Area under the curve: look at the units for the area under the curve to see if it corresponds to a measurement.

Graphical Representation of Motion
Physics C 1-D Motion Graphical Representation of Motion Position-Time Graph A B x Dx Dt slope t Steepness = speed slope Sign = direction Velocity = speed + direction Bertrand

Average Velocity from a Graph
Physics C 1-D Motion Average Velocity from a Graph t x A B Dx Dt Mathematical Model x = position x0 = initial position v = average velocity t = time Bertrand

Graphs of Motion Mathematical Model x0 = 20 m Slope v = 2 m/s
Physics C 1-D Motion Graphs of Motion Mathematical Model Dx =20m Dx =20m x0 = 20 m v = 2 m/s Slope SLOPE AREA Mathematical Model (UNIFORM VELOCITY) Area=20m = Dx Area (init and final positions unknown. ONLY KNOW DISPLACEMENT) Bertrand 22

What is happening in this graph?
Mathematical Graphical Verbal: For every second that passes the buggy travels +0.4 meter every second from an initial position of 0.5 meter in the positive x direction. NEGATIVE SPEED??? End t=6 s 5s 4s 3s 2s 1s t=0 s START X = -10 -5 5 10 15 20 m Diagrammatic v = 23

Plot the corresponding v-t graph
x0 = 20 m v = -5 m/s Slope SLOPE 24

What is happening? Follow pos – goes away from starting point then at C comes closer Always at same speed? Moving away from starting point to A Slows down to B Speeds up to C Reverses direction and comes closer to start with

What is happening in each?
D. B.

Draw the corresponding v-t graph v-t
Physics C 1-D Motion Draw the corresponding v-t graph SLOPE v-t Bertrand

F=D=E>A>B>C C
Physics C 1-D Motion B A Which regions shows positive displacement? negative? When is the object moving in the + direction? When is the object moving in the – direction? Which region is the object moving with maximum + velocity? Rank speeds from greatest to least When is the object at rest When does the object change direction? A,F B,D,E A,F B,D,E F F=D=E>A>B>C C At 2s, at 7 s Bertrand

Physics C 1-D Motion B A What is the total displacement for the 8 sec? What is the average velocity for the entire 8 sec trip? What is the distance traveled in the 8 sec trip? What is the average speed for the 8 sec trip? 0m 0m/s 32m 4m/s Bertrand

Problem: A car starting from rest moves with an average speed of 6 m/s
Problem: A car starting from rest moves with an average speed of 6 m/s. Calculate the distance the car traveled in 1 minute. GIVEN: s = 6 m/s t = 1 min =60 s UNKNOWN: d = ? m FORMULA: s = d / t SUBSTITUTION: 6 m/s = d / 60 s SOLUTION d = 360 m

Problem: An object moves from the position +47 m to the position +15 m in 8 s. What is its total displacement? What is its average velocity? GIVEN: xo = 47m xf = 15m t = 8 s UNKNOWN: Dx = ? m FORMULA: Dx=xf-xo SUBSTITUTION: Dx= 15-47 SOLUTION Dx = -32 m GIVEN: xo = 47m xf = 15m Dx = -32 m t = 8 s UNKNOWN: vav = ? m/s FORMULA: SUBSTITUTION: SOLUTION

v-t Draw the corresponding x-t graph
Physics C 1-D Motion AREA v-t Dx= +12m Draw the corresponding x-t graph 3m Area= Dx -6m Dx= -16m Bertrand

The Meaning of Shape of a Position-Time graph
Contrast a constant and changing velocity Contrast a slow and fast moving object

Average Velocity and Instantaneous Velocity
Average velocity only depends on the initial and final positions. These 2 cars have the same average velocities but different velocities at each instant. When the velocity is not uniform, the instantaneous velocity is not the same as the average velocity. AVERAGE VELOCITY: X = -40 -20 20 40 60 80 km X = -40 -20 20 40 60 80 km Start t=0 End t=2 hr

Physics C 1-D Motion Acceleration Average acceleration describes how quickly or slowly the velocity changes. It is calculated by: Vector SI units: m/s2 Instantaneous acceleration describes how the velocity changes over a very short time interval: Acceleration tells us how fast the velocity changes. Velocity tells us how fast the position changes. Bertrand

x v a Average acceleration
Example - A car accelerates along a straight road from rest to 24 m/s in 6.0 s. What is the average acceleration? Average acceleration START 1s 2s 3s 4s 5s x v 4 8 12 16 20m/s a 4m/s2

x x v a Position-Time Graph t=0 t=5 t=4 t=3 t=2 t=1 1s 2s 3s 4s 5s
START 1s 2s 3s 4s 5s x

x v a x v a Speeding up in + direction a and v SAME direction
START 1s 2s 3s 4s 5s x v a Slowing down in + direction a and v OPP direction START 1s 2s 3s 4s 5s x v a

x v a x v a Speeding up in - direction a and v SAME direction
START 5s 4s 3s 2s 1s x v a Slowing up in - direction a and v OPP direction START 5s 4s 3s 2s 1s x v a

Displacement and velocity are in the direction of motion
Physics C 1-D Motion Displacement and velocity are in the direction of motion When acceleration is in the SAME direction as velocity, the object is speeding up When acceleration is in the OPPOSITE direction to velocity, the object is slowing down Bertrand

x x v a Speeding up in + direction t=0 t=5 t=4 t=3 t=2 t=1 1s 2s 3s 4s
Position-Time Graph START 1s 2s 3s 4s 5s x

x Slowing down in + direction x v a t=0 t=5 t=3 t=2 t=1 t=4 1s 2s 3s
Position-Time Graph START 1s 2s 3s 4s 5s x

x Speeding up in - direction x v a t=5 t=0 t=1 t=2 t=3 t=4 5s 4s 3s 2s
START 5s 4s 3s 2s 1s x

x Slowing down in - direction x v a t=0 t=1 t=2 t=3 t=4 t=5 5s 4s 3s
START 5s 4s 3s 2s 1s x

x What is this object doing? Slowing down in - direction x v a t=5 t=0
START 5s 4s 3s 2s 1s x

Draw the corresponding v-t and a-t graphs x-t graph v-t graph

x What is this object doing? Speeding up in - direction x v a t=5 t=0
START 5s 4s 3s 2s 1s x

PHYSICS DEPARTMENT STORE
SLOPE AREA SLOPE AREA

Constant Acceleration Motion
AVERAGE VELOCITY SLOPE = acceleration Connect with curved line No physical meaning AREA slope AVERAGE acceleration SLOPE = AREA AREA slope AREA SLOPE = Dv =-9 AREA RUNNING TOTAL

constant acceleration
SLOPE = Velocity NOT constant slope AREA SLOPE = Dx AREA slope constant acceleration AREA AREA Dv

Estimate the net displacement from 0 s to 5.0 s
Physics C 1-D Motion Estimate the net displacement from 0 s to 5.0 s Area under v-t curve Dx = = 10.5 m Bertrand

Construct the corresponding x-t and a-t curves
Physics C 1-D Motion Construct the corresponding x-t and a-t curves Straight (constant v) Curved (acceleration) AREA slope Bertrand 52

Estimate the displacement from 0 s to 4.0 s
Physics C 1-D Motion Estimate the displacement from 0 s to 4.0 s Area under v-t curve Dx = 2 -2 = 0 m Bertrand

Construct the corresponding x-t and a-t curves
Physics C 1-D Motion All Curved (acceleration) AREA slope Bertrand 54

Construct the corresponding x-t and a-t curves
curved Construct the corresponding x-t and a-t curves straight straight curved curved B AREA A C What region moving + direction, -direction, reverse? What region speeding up, slowing down, constant v, stopped D F E slope 55

Representations of Accelerated Motion Graphical Mathematical
x-t graph is quadratic: A and B have physical meaning which we will derive v-t graph is linear: a-t graph a = constant

(Equation of v-t graph)
Mathematical Equations to Represent Constant Accelerated Motion and relationship to graphs Definition of average velocity: (Slope of x-t graph) 1 Definition of average acceleration: 2 (Equation of v-t graph) (Slope of v-t graph) constant acceleration For constant acceleration: 3

Mathematical Equations to Represent Constant Accelerated Motion and relationship to graphs
1 4

(Constant Acceleration)
KINEMATIC EQUATIONS (Constant Acceleration) Dx = vDt (definition of average velocity) Dv = aDt (definition of avr a) Dt = (vf – v0)/a v v0 + vf 2 = (average velocity for constant acceleration) Dx = (v0 + vf)(vf – v0)/2a 5. vf2 = v02 + 2aDx (time independent)

(Constant Acceleration)
KINEMATIC EQUATIONS (Constant Acceleration) 2. 3. 4. 5.

Representations of Accelerated Motion Graphical Mathematical
x-t graph is quadratic: t x = At2 +Bt + x0 x = ½at2 +v0t + x0 v-t graph is linear: a-t graph a = constant

Do Now (solve graphically): A car moving at 12 m/s is 36 m away from a stop sign. What acceleration will stop the car exactly at the stop sign? v-t 12 Slope=a a = -2 m/s2 Solve graphically Area Dx=36m tstop

Example: A car moving at 12 m/s is 36 m away from a stop sign
Example: A car moving at 12 m/s is 36 m away from a stop sign. What acceleration will stop the car exactly at the stop sign? v0= 12 m/s vf = 0 a = ? m/s2 Dx = 36 m t a = -2 m/s2

5-Step Problem Solving GIVEN UNKNOWN FORMULA SUBSTITUTION SOLUTION
Read & Draw Diagram (when needed) List all the given information using variables and units NO WORDS UNKNOWN What variable are you looking to solve Variable = ? units FORMULA Write formula SUBSTITUTION Substitute givens into formula (include units) SOLUTION Box your final answer: make sure units are expressed

Dx = 816 m KINEMATICS PROBLEMS
A passenger jet lands on a runway with a velocity of 70 m/s. Once it touches down, it accelerates at a constant rate of -3 m/s2. How far does the plane travel down the runway before its velocity is decreased to 2 m/s, its taxi speed to the landing gate? vi = 70 m/s vf = 2 a = -3 m/s2 Dx = ? m t Dx = 816 m a a vi vf Dx

vi = 1.75 m/s vf = 6.25 m/s KINEMATICS PROBLEMS
A runner goes 12 m in 3 s at a constant acceleration of 1.5 m/s2. What is her velocity at the end of the 12 m? vi = vf = ? m/s a = 1.5 m/s2 Dx = 12 m t = 3 s vi = 1.75 m/s vf = 6.25 m/s

vf = 3 m/s KINEMATICS PROBLEMS
The U.S. and South Korean soccer teams are playing in the first round of the world cup. An American kicks the ball giving it an initial velocity of 4 m/s. The ball rolls a distance of 7 m and is then intercepted by a South Korean player. If the ball accelerated at m/s2 while rolling across the grass, find its velocity at the time of interception. vi = 4 m/s vf = ? m/s a = m/s2 Dx = 7 m t = Signs and root issue vf = 3 m/s