1 – D Motion• We begin our discussion of kinematics (description

of motion in mechanics)– Simplest case: motion of a particle in 1 – D– Concept of a particle: idealization of treating a body as a

single point (we get close to doing this by pinpointing the “license plate of a car,” etc.)

– 1 – D motion: Only two possible directions

• We will be working with vector quantities– Characterized by both a magnitude AND a direction– Scalar quantities have a magnitude but NOT a direction

• How do we describe motion in physics?• Should be able to answer questions, “Where is it?”

along with “Where is it going?” and “How fast is it going?”

Position and Displacement• Example: Motion of a car along Sandusky St.

Park Ave. (reference

point or origin)

Central Ave. (point 2)

William St. (point 1)

+x

2x

1x

:, 21 xx

Position vectors (keep track of position of car)

:12 xx

Displacement vector (change in position from point 1 to point 2)

12 xx

xixx

ixix

ˆ)blocks4(

ˆ)blocks6(andˆ)blocks2(For

12

21

(north)

CQ1: Consider the figure below that shows three paths between position 1 and position 2. (Path C is a half

circle.) Which path would result in the greatest displacement for a particle moving from position 1 to

position 2?

A) Path A

B) Path B

C) Path C

D) All paths would result in the same displacement.

• Velocity describes speed (magnitude) and direction– Need time as well as position information

• Average velocity:

Velocity

Park Ave. (reference

point or origin)

Central Ave. (point 2)William St. (point 1)

+x

Stopwatch measures time t1

Stopwatch measures time t2

t

x

tt

xxv

12

12ave

(average speed = total distance traveled / time to travel that distance)

2x

1x

Velocity• Average velocity could be positive or negative

– Trip from William St. to Central Ave. (previous picture) – assume t1 = 0 s and t2 = 40 s:

– Trip from Central Ave. to William St.:

ii

tt

xxv ˆ)s/blocks1.0(

)s0s40(

ˆ)blocks4(

12

12ave

Park Ave. (origin)

Central Ave. (point 1) William St. (point 2)

+x

Stopwatch measures time t2

Stopwatch measures time t1

1x2x

12 xx

Velocity– Assume t1 = 0 and t2 = 40 s:

• Ave. velocity depends only on the total displacement that occurs during time interval t = t2 – t1, not on what happens during t:

ii

tt

xxv ˆ)s/blocks1.0(

)s0s40(

ˆ)blocks4(

12

12ave

x

timePark Ave.

William St.

Central Ave. Trip #2

vave is the same for both trips!

t1 t2

Trip #1

Velocity– Note that each line represents how position changes with

time and is not representative of a path in space– Slope of line between 2 points on this graph gives average

velocity between these points

• Instantaneous velocity = velocity at any specific instant in time or specific point along path– Mathematically, it is the limit as t 0 of the average

velocity:

– Say we wanted to know velocity at time t1

– Imagine moving t2 closer and closer to t1

– x & t become very small, but ratio not necessarily small

t

xv

t

0lim

Instantaneous Velocity• Graphically, instantaneous velocity at a point =

slope of line tangent to curve at that point

x

timet2

Slope = instant. velocity at time t2

t1

Slope = instant. velocity at time t1

Park Ave.

William St.

Central Ave.

Velocity• Be careful not to interchange “speed” with

“velocity”– “speed” refers to a magnitude only– “velocity” refers to magnitude and direction– For example, I might run completely around a

1–mi. circular track in ¼ hour, with an average speed (round-trip) = dist. traveled / time = 1 mi / 0.25 hr = 4 mph

– BUT since

– In general: instantaneous speed = but average speed

0ave

t

xv

0x

(I’m right back where I started!)

v

avev

CQ2: The graph below represents a particle moving along a line. What is the total distance traveled by the

particle from t = 0 to t = 10 seconds?

A) 0 m

B) 50 m

C) 100 m

D) 200 m

Acceleration• Acceleration describes changes in velocity

– Rate of change of velocity with time– Vector quantity (like velocity and displacement)– “Velocity of the velocity”

• Average acceleration defined as:– are the instantaneous velocities at t2 and t1,

respectively

• To find average acceleration graphically:

t

v

tt

vva

12

12ave

12 ,vv

v

t

v1

v2

t1 t2

Slope = average acceleration

Acceleration• We can also find the acceleration at a

specific instant in time, similar to velocity

• Instantaneous acceleration:

• Graphically:

• BE CAREFUL with algebraic sign of acceleration – negative sign does not always mean that body is slowing down

t

va

t

0lim

v

t

v1

t1

Slope of tangent line = instantaneous acceleration

Acceleration– Example: Car has neg. acceleration if slowing down from

pos. velocity, or if it’s speeding up in neg. direction (going in reverse)

– Must compare direction of velocity and acceleration:

– Be careful not to interchange deceleration with negative acceleration

v a motion

+ + Speeding up

– – Speeding up

+ – Slowing down

– + Slowing down(for example, going slower in reverse)

Same sign

Opposite sign

CQ3: Which animation shown in class gave the correct position vs. time graph? (Assume that the

positive x direction is to the right.)

A) Animation 1

B) Animation 2

C) Animation 3

D) Animation 4

PHYSLET #7.1.1, Prentice Hall (2001)

Constant Acceleration• For special case of constant acceleration,

then v vs. t graph becomes a straight line

– Note that slope of above line (= acceleration) is the same for any line determined from 2 points on the line or for any tangent line since (for this special case)

– We can find the instantaneous acceleration from:

v

t

Constant acceleration means constant rate of increase for v

aa

ave

ave12

12 att

vva

Constant Acceleration• Now let t1 = 0 (can start our clock whenever

we want) and t2 = t (some arbitrary time later), and define v(t1= 0) = v0 and v(t2= t) = v

• Graphical interpretation:

ave0

0a

t

vva

)1(

0 tavv

(takes form of equation of a line: y = b + mx)

v

tv0

v

at

v0

Average value of v: )2(2

0ave

vvv

(since v is a linear function)

Also, by definition: )3(0

12

12ave t

xx

tt

xxv

Setting (2) = (3), and using (1), we get:

)4(2

1 200 tatvxx

Constant Acceleration• We can use equations (1) and (4) together to

obtain an equation independent of t and another equation independent of a using substitution, with the result:

tvv

xx

xxavv

2

)(2

00

020

2 (independent of t)

(independent of a)

Constant Acceleration• Graphs of a, v, and x versus time:

Motion Graphs Interactive

CQ4: The graph below represents a particle moving along a line. When t = 0, the displacement of the particle

is 0. All of the following statements are true about the particle EXCEPT:

A) The particle has a total displacement of 100 m.

B) The particle moves with constant acceleration from 0 to 5 s.

C) The particle moves with constant velocity between 5 and 10s.

D) The particle is moving backwards between 10 and 15 s.

Interactive Example Problem:Seat Belts Save Lives!

Animation and solution details given in class.

ActivPhysics Problem #1.8, Pearson/Addison Wesley (1995–2007)

Example Problem #2.39

Solution (details given in class):

(a) 1.5 m/s

(b) 32 m

A car starts from rest and travels for 5.0 s with a uniform acceleration of +1.5 m/s2. The driver then applies the brakes, causing a uniform acceleration of –2.0 m/s2. If the brakes are applied for 3.0 s,

(a) how fast is the car going at the end of the braking period, and

(b) how far has the car gone?

Example Problem #2.43

Solution (details given in class):

(a) 8.2 s

(b) 134 m

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 12 m/s, skates by with the puck. After 3.0 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 4.0 m/s2,

(a) how long does it take him to catch his opponent, and

(b) how far has he traveled in that time? (Assume that the player with the puck remains in motion at constant speed.)

Free Fall• Particular case of motion with constant acceleration:

“free-fall” due to gravity– Objects falling to the ground– Galileo (16th Century) showed that bodies fall with

constant downward acceleration, independent of mass (neglecting air resistance and buoyancy effects)

• Near the Earth’s surface, this acceleration has a constant value of g 9.8 m/s2

• We can use all of the constant-acceleration equations for free fall– Will use “y” instead of “x” when referring to vertical

positions:2

00 2

1tatvyy

Free Fall• Sign of depends on choice of +y – axis direction:

– Choice of axis not important – as long as you are consistent!

– always points toward the ground, however

a

0

+y (up)

then = a

g

then = a

g

0

+y (down)

a

a

a

CQ5: A ball is thrown straight up in the air. For which situation are both the instantaneous

velocity and the acceleration zero?

A) on the way up

B) at the top of the flight path

C) on the way down

D) halfway up and halfway down

E) none of these

Free Fall Example ProblemHow long would it take for Tom Petty to go “Free-Fallin’” from the top of the Sears Tower in Chicago? (Of course there is a safety net at the bottom.)

+y

443 m

y = y0 + v0t – ½ gt2

– 443 m = 0 + 0 – ½ gt2

886 m / g = t2 t2 = 90.4 s2

t = 9.5 s t = +9.5 s (neg. value has no physical meaning)How fast will he be moving just before he hits the net?

v = v0 – gt = 0 – gt = – (9.8 m/s2)(9.5 s) = – 93.1 m/s

(negative sign means downward direction)

CQ6: A pebble is dropped from rest from the top of a tall cliff and falls 4.9 m after 1.0 s has

elapsed. How much farther does it drop in the next 2.0 seconds?

A) 9.8 m

B) 19.6 m

C) 39 m

D) 44 m

E) 27 m

Final Comments About 1–D Motion• In reality, Tom Petty’s velocity would “tail off” and

start reaching a maximum:

• Other notes about 1–D velocity and acceleration:– In general, both velocity and acceleration are not constant

in time– Can use methods of calculus to generalize equations – General formulas reduce to constant acceleration formulas

when a(t) = constant = a– Remember that formula d = vt only holds when

v = constant!

v

t

vt

Free fall only

Including air resistance effects

vt is called the “terminal velocity”