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Physics 101Lecture 3
Motion in 1D
Assist. Prof. Dr. Ali ÖVGÜN
EMU Physics Department
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Jan. 28-Feb. 1, 2013
Motion along a straight lineq Motionq Position and displacementq Average velocity and average speedq Instantaneous velocity and speedq Acceleration q Constant acceleration: A special
caseq Free fall acceleration
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MotionqEverything moves!
Motion is one of the main topics in Physics I
qSimplification: Consider a moving object as a particle, i.e. it moves like a particle—a “point object”
LAX
Newark
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4 Basic Quantities in Kinematics
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One Dimensional Position rq Motion can be defined as the change of position over
time.q How can we represent position along a straight line?q Position definition:
n Defines a starting point: origin (r = 0), r relative to originn Direction: positive (right or up), negative (left or down)n It depends on time: t = 0 (start clock), r(t=0) does not have to
be zero.q Position has units of [Length]: meters.
r = + 2.5 m i r = - 3 m i
For motion along a straight line, the direction is represented
simply by + and – signs.+ sign: Right or Up.- sign: Left or Down.
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Displacementq Displacement is a change of position in time.q Displacement:
n f stands for final and i stands for initial.q It is a vector quantity.q It has both magnitude and direction: + or - signq It has units of [length]: meters.
)()( iiff trtrr !!!-=D
r1 (t1) = + 2.5 m ir2 (t2) = - 2.0 m i
Δr = -2.0 m - 2.5 m = -4.5 m ir1 (t1) = - 3.0 m ir2 (t2) = + 1.0 m i
Δr = +1.0 m + 3.0 m = +4.0 m i
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Distance and Position-time graph
q Displacement in spacen From A to B: Δr = rB – rA = 52 m – 30 m i = 22 m in From A to C: Δr = rc – rA = 38 m – 30 m = 8 m i
q Distance is the length of a path followed by a particlen from A to B: d = |rB – rA| = |52 m – 30 m| = 22 mn from A to C: d = |rB – rA|+ |rC – rB| = 22 m + |38 m – 52 m| = 36 m
q Displacement is not Distance. www.aovgun.com
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Velocityq Velocity is the rate of change of position.q Velocity is a vector quantity.q Velocity has both magnitude and direction.q Velocity has a unit of [length/time]: meter/second.q We will be concerned with three quantities, defined as:
n Average velocity
n Average speed
n Instantaneous velocity
avgtotal distances
t=
D
0limt
x dxvt dtD ®
D= =
D
txx
txv if
avg D
-=
DD
=
displacement
distance
displacement
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Average Velocityq Average velocity
is the slope of the line segment between end points on a graph.
q Dimensions: length/time (L/T) [m/s].
q SI unit: m/s.q It is a vector (i.e. is signed), and
displacement direction sets its sign.
txx
txv if
avg D
-=
DD
=
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Average Speedq Average speed
q Dimension: length/time, [m/s].q Scalar: No direction involved.q Not necessarily close to Vavg:
n Savg = (6m + 6m)/(3s+3s) = 2 m/sn Vavg = (0 m)/(3s+3s) = 0 m/s
avgtotal distances
t=
D
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Instantaneous Velocityq Instantaneous means “at some given instant”. The
instantaneous velocity indicates what is happening at every point of time.
q Limiting process:n Chords approach the tangent as Δt => 0n Slope measure rate of change of position
q Instantaneous velocity:q It is a vector quantity.q Dimension: length/time (L/T), [m/s].q It is the slope of the tangent line to x(t).q Instantaneous velocity v(t) is a function of time.
0limt
x dxvt dtD ®
D= =
D
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q Uniform velocity is the special case of constant velocityq In this case, instantaneous velocities are always the
same, all the instantaneous velocities will also equal the average velocity
q Begin with then
Uniform Velocity
txx
txv if
x D
-=
DD
= tvxx xif D+=
xx(t)
t0xi
xf
v
v(t)
t0tf
vx
ti
Note: we are plotting velocity vs. time
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Average Accelerationq Changing velocity (non-uniform) means an
acceleration is present.q Acceleration is the rate of change of velocity.q Acceleration is a vector quantity.q Acceleration has both magnitude and direction.q Acceleration has a dimensions of length/time2: [m/s2].q Definition:
n Average acceleration
n Instantaneous accelerationif
ifavg tt
vvtva
-
-=
DD
=
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2
0lim dt
xddtdx
dtd
dtdv
tva
t===
DD
=®D
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Average Accelerationq Average acceleration
q Velocity as a function of time
q It is tempting to call a negative acceleration a “deceleration,” but note:n When the sign of the velocity and the acceleration are the
same (either positive or negative), then the speed is increasing
n When the sign of the velocity and the acceleration are in the opposite directions, the speed is decreasing
q Average acceleration is the slope of the line connecting the initial and final velocities on a velocity-time graph
if
ifavg tt
vvtva
-
-=
DD
=
tavtv avgif D+=)(
Note: we are plotting velocity vs. time
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Instantaneous and Uniform Acceleration
q The limit of the average acceleration as the time interval goes to zero
q When the instantaneous accelerations are always the same, the acceleration will be uniform. The instantaneous acceleration will be equal to the average acceleration
q Instantaneous acceleration is the slope of the tangent to the curve of the velocity-time graph
2
2
0lim dt
xddtdx
dtd
dtdv
tva
t===
DD
=®D
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Special Case: Motion with Uniform Acceleration (our typical case)
q Acceleration is a constantq Kinematic Equations (which
we will derive in a moment)
tavv !!!+= 0
221
00 tatvrr !!!!++=
xavv !!!!D+= 22
02
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Problem-Solving Hintsq Read the problemq Draw a diagram
n Choose a coordinate system, label initial and final points, indicate a positive direction for velocities and accelerations
q Label all quantities, be sure all the units are consistentn Convert if necessary
q Choose the appropriate kinematic equationq Solve for the unknowns
n You may have to solve two equations for two unknownsq Check your results
xavv D+= 220
2
atvv += 0
221
0 attvx +=D
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Exampleq An airplane has a lift-off speed of 30 m/s
after a take-off run of 300 m, what minimum constant acceleration?
q What is the corresponding take-off time?
xavv D+= 220
2
atvv += 02
21
0 attvx +=D xavv D+= 220
2
atvv += 02
21
0 attvx +=D
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Free Fall Accelerationq Earth gravity provides a constant
acceleration. Most important case of constant acceleration.
q Free-fall acceleration is independent of mass.
q Magnitude: |a| = g = 9.8 m/s2
q Direction: always downward, so ag is negative if we define “up” as positive,a = -g = -9.8 m/s2
q Try to pick origin so that xi = 0
y
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q A stone is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The stone just misses the edge of the roof on the its way down. Determine
q (a) the time needed for the stone to reach its maximum height.
q (b) the maximum height.q (c) the time needed for the stone to return to the height
from which it was thrown and the velocity of the stone at that instant.
q (d) the time needed for the stone to reach the groundq (e) the velocity and position of the stone at t = 5.00s
Falling for Rookie
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Example
Example
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Summaryq This is the simplest type of motionq It lays the groundwork for more complex motionq Kinematic variables in one dimension
n Position r(t) m Ln Velocity v(t) m/s L/Tn Acceleration a(t) m/s2 L/T2
n All depend on timen All are vectors: magnitude and direction vector:
q Equations for motion with constant acceleration: missing quantitiesn r
n v
n t
tavv !!!+= 0
221
00 tatvrr !!!!++=
)(2 020
2 xxavv !!!!"-+=
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Problem2
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Problem1
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Problem3
Problem4
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Problem5
Problem6
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Problem7
Problem8
Problem9
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Problem10
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24.5 m/s and 0.1 m/s
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