Physics 111Lecture 3Motion in

One DimensionDr. Ali ÖVGÜN

EMU Physics Department

www.aovgun.com

Jan. 28-Feb. 1, 2013

Jan. 28-Feb. 1, 2013

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

Jan. 28-Feb. 1, 2013

4 Basic Quantities in Kinematics

Jan. 28-Feb. 1, 2013

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.

Jan. 28-Feb. 1, 2013

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 !!! −=Δ

r1 (t1) = + 2.5 m ir2 (t2) = - 2.0 m iΔr = -2.0 m - 2.5 m = -4.5 m ir1 (t1) = - 3.0 m ir2 (t2) = + 1.0 m iΔr = +1.0 m + 3.0 m = +4.0 m i

Jan. 28-Feb. 1, 2013

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.

Jan. 28-Feb. 1, 2013

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 avgtotal distances

t=

Δ

txx

txv if

avg Δ−

=ΔΔ=

displacement

distance

Jan. 28-Feb. 1, 2013

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 acceleration if

ifavg tt

vvtva

−−

=ΔΔ=

Jan. 28-Feb. 1, 2013

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 !!!! Δ+= 220

2

Jan. 28-Feb. 1, 2013

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 Δ+= 220

2

atvv += 0

221

0 attvx +=Δ

Jan. 28-Feb. 1, 2013

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 Δ+= 220

2

atvv += 02

21

0 attvx +=Δ xavv Δ+= 220

2

atvv += 02

21

0 attvx +=Δ

Jan. 28-Feb. 1, 2013

September 8, 2008

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 define “up” as positive,a = -g = -9.8 m/s2

q Try to pick origin so that yi = 0

y

September 8, 2008

Free Fall Accelerationq Two important equation:

q Begin with t0 = 0, v0 = 0, y0 = 0q So, t2 = 2|y|/g same for two balls!q Assuming the leaning tower of Pisa is

150 ft high, neglecting air resistance,t = (2´150´0.305/9.8)1/2 = 3.05 s

y

gtvv −= 0

20 0

12

y y v t gt= + −0

Jan. 28-Feb. 1, 2013

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 !!!!" −+=

September 8, 2008

September 8, 2008

September 8, 2008

September 8, 2008

September 8, 2008

September 8, 2008

24.5 m/s and 0.1 m/s

September 8, 2008

P8: