Pre AP Physics

One Dimensional Motion

Velocity, Acceleration, Motion Graphs

Velocity

● Displacement divided by time interval● Velocity is the change in position divided by the

time interval during which the motion took place.● Velocity is a vector quantity

● Speed is scalar and is distance divided by time.

Types of Velocity● Average Velocity – the total displacement of motion

divided by the total time interval.● Instantaneous Velocity – the velocity in a specific

instant. Like a quick check on your speedometer.● Initial Velocity – the velocity at the start of the time

interval. When object starts at rest, Vi = zero.● Final Velocity – the velocity of the object at the end of

the time interval. For an object braking to a stop, Vf = zero.

● Terminal Velocity – got to wait for the force chapter, sorry.

Average Velocity Equation

Vavg=

Manipulate to solve for d and t

Velocity and Direction

Practice Problems 2A p.44 1-5

ClassWork

1. How far have you walked if your velocity is 2m/s West and you walk for 2 minutes?

2. How much time would it take to drive 2km if your average velocity is 30m/s?

3. What is Mr. Crabtree’s average velocity if he runs 3.2km in a time of 18 minutes and 20 seconds?

Distance – Time GraphsAlso called Position – Time Graphs

Position (m)

Time (sec)

Slope – what does it mean?

Constant Velocity

● A special Case● An unchanging Velocity● Not speeding up, not slowing down, not changing direction.

● Its like perfect cruise control on a level, straight highway.

Change in VelocityChange in Velocityspeed up, slow down, or change in speed up, slow down, or change in

directiondirection

ā= V2-V1

t2-t1

Units= m/s s m/s2 ms-2

Acceleration is a vector Acceleration is a vector quantity (like velocity)quantity (like velocity)Direction makes a Direction makes a difference!difference!

If you have If you have (+) positive velocity(+) positive velocity and and speed speed upup = =(+)positive (+)positive accelerationacceleration (+) positive velocity(+) positive velocity and and slow slow downdown = =(-)negative (-)negative accelerationacceleration

An acceleration value tells you An acceleration value tells you by how by how muchmuch an object is changing its an object is changing its velocityvelocity

Time Time VelocityVelocity

0 sec O 0 sec O m/sm/s

1 sec 3 1 sec 3 m/sm/s

2 sec 6 2 sec 6 m/sm/s

3 sec 9 3 sec 9 m/sm/s

4 sec 4 sec 12 m/s12 m/s

5 sec 5 sec 15 m/s15 m/s

For an For an object object

acceleratiaccelerating @ ng @ 3m/s3m/s22

Example ProblemExample Problem The velocity of a car increases from The velocity of a car increases from 2 m/s2 m/s at at 1 sec.1 sec.

To To 16 m/s at 4.5 sec.16 m/s at 4.5 sec. What is the car’s What is the car’s average average accelerationacceleration??

Given Formula SolutionGiven Formula Solution

VVii= 2m/s a= v a= 16m/s – 2 m/s= 2m/s a= v a= 16m/s – 2 m/s

VVff= 16m/s t 4.5 sec- 1 sec= 16m/s t 4.5 sec- 1 sec

TT11= 1 sec a= 14m/s= 1 sec a= 14m/s

TT22= 4.5 sec 3.5 sec = 4.5 sec 3.5 sec = 4m/s4m/s22

Look at the following Look at the following Practice ProblemsPractice Problems

p.49p.491,2,31,2,3

p.49p.49SampleSample

On A velocity-TimeOn A velocity-TimeGraph: Graph:

Slope= AccelerationSlope= Acceleration

Sample 2B p.49● A shuttle bus slows to a stop with an average

acceleration of –1.8m/s2. How long does it take the bus to slow from 9m/s to 0m/s?

Practice● What is the acceleration of a car that goes from 4m/s to

36m/s in a time interval of 4 seconds?

● What is the acceleration of a car that goes from 36m/s to 15m/s in a time interval of 3 seconds?

● What is the acceleration of a vehicle that goes from –3m/s to 4.5m/s in a time of 2.5 seconds?

● Solutions on the next 2 slides

#1#1. . GivenGiven FormulaFormula SolutionSolution

VVii== 4m/s a 4m/s a == V Vff-V-Vii

VVff== 36m/s t 36m/s t

tt== 4s 36 – 4 4s 36 – 4

44

#2. V#2. Vii== 36 m/s a 36 m/s a == VVff--VVi i

VVff== 15 m/s t 15 m/s t

tt== 3 sec 15 3 sec 15 – – 3636

33

8 m/s2

- 7 m/s- 7 m/s22

#3. V#3. Vii= -3 m/s a= V= -3 m/s a= Vff – V – Vii

VVff= + 4.5 m/s t= + 4.5 m/s t

t= 2.5 sec 4.5 – (-3)t= 2.5 sec 4.5 – (-3)

2.52.5

3 m/s2

GREAT JOB!

Acceleration that does not change is called Acceleration that does not change is called uniform or constant accelerationuniform or constant acceleration

(for simplicity sake, the problems we will solve (for simplicity sake, the problems we will solve will be in uniform acceleration)will be in uniform acceleration)

22ndnd Formula– Formula–

VVff= V= Vii ++ atat

VVii= V= Vff – at t = V – at t = V ff – V – Vii a= V a= Vff – V – Vi i

a a t t

● If a car with a velocity of 2 m/s accelerates at a rate of (+)4 m/s2 for 2.5 sec., what is the velocity at 2.5 sec?

Given Formula Solution

Vi= 2 m/s Vf= Vi + at

A = 4 m/s2 Vf= 2 + 4(2.5)

T= 2.5 sec

Vf= ??

Vf = 12 m/s

Do sample problem p. 55

Sample Problem p.55● A plane starting from rest at one end of the runway undergoes a uniform

acceleration of 4.8m/s2 for 15 seconds before takeoff. What is the speed at takeoff?

Displacement Displacement when when velocityvelocity and and timetime are knownare known

d= ½ (Vd= ½ (Vff + V + Vii) t ) t

**True when object is accelerated uniformlyTrue when object is accelerated uniformly Displacement when Displacement when accelerationacceleration and and

timetime are known are known

d= Vd= Viit + ½ att + ½ at22

**recall that an object starting from rest has Vrecall that an object starting from rest has Vii= = 0 m/s0 m/s

When starting from When starting from restrest

d= ½ atd= ½ at22

Sample Problem● A plane starting from rest at one end of the runway undergoes a uniform

acceleration of 4.8m/s2 for 15 seconds before takeoff. How long must the runway be for the plane to be able to take off?

2 C

Page 53 practice problems 1-4

2 D

Page 55 practice problems 1-4

● A certain airplane wants to takeoff from A certain airplane wants to takeoff from a runway that is 1 kilometer long. The a runway that is 1 kilometer long. The plane starts from rest and must get to a plane starts from rest and must get to a speed of 71m/s in order to lift off the speed of 71m/s in order to lift off the ground. What is the minimum ground. What is the minimum acceleration the pilot has to achieve to acceleration the pilot has to achieve to order to takeoff?order to takeoff?

** you are not given time.** you are not given time.

● Answer is on the next slide.Answer is on the next slide.

Solving without timeSolving without time

VVff22= V= Vii

22 + 2 ad + 2 ad

VVff= V= Vii22+ 2ad a= V+ 2ad a= Vff

22 – V – Vii22 d= V d= Vff

22 – V – Vii22

2d 2a2d 2a

Given Formula SolutionGiven Formula Solution

VVii= 0 a= V= 0 a= Vff22- V- Vii

22 a = 71 a = 7122 - 0 - 0

VVff= 71 m/s 2d 2(1000) = 71 m/s 2d 2(1000)

d= 1000md= 1000m

2.5 m/s2.5 m/s22

HW pages HW pages

2 E

Page 58 practice problems 1-4

Velocity Time Graphs

Velocity (m/s) Time

(sec)

V – t Graphs

● Slope equals ______________________● Area under the slope (curve, line) equals

________________

examples

Velocity (m/s)

Time (sec)

Class Work

Page 68 N and P

D-t and V-t graphs side by side.

Acceleration due to Acceleration due to gravitygravity

Earth’s gravity pulls everything towards Earth’s gravity pulls everything towards the center of the earththe center of the earth

All things are pulled at the same rate. All things are pulled at the same rate. (atoms to elephants- all the same) – so long (atoms to elephants- all the same) – so long as we neglect air resistance, which we will as we neglect air resistance, which we will most of the time. most of the time.

Gravity is a uniform acceleration Gravity is a uniform acceleration Gravity on Earth can change as you go from Gravity on Earth can change as you go from

place to place. Gravity changes as you place to place. Gravity changes as you change the distance from the center of the change the distance from the center of the earth to your location. earth to your location.

EXAMPLE: on a mountain gravity is less EXAMPLE: on a mountain gravity is less than at sea level. also earth isn’t perfectly than at sea level. also earth isn’t perfectly round gravity is less at equator than polesround gravity is less at equator than poles

Gravity can vary from place to place; but Gravity can vary from place to place; but only at the same place are all objects only at the same place are all objects accelerated the same. accelerated the same.

Every planet- (as well as the moon and Every planet- (as well as the moon and sun) has its own acceleration due to sun) has its own acceleration due to gravity gravity

size and mass determine gravitysize and mass determine gravity acceleration due to gravity is a vector acceleration due to gravity is a vector

quantityquantity

● We will use an average value for gravity We will use an average value for gravity when solving problems on Earth.when solving problems on Earth.

g= - 9.81 m/sg= - 9.81 m/s22

● As an object falls w/out air on EarthAs an object falls w/out air on Earth

Time VelocityTime Velocity

0s 0 0s 0

1s -9.8 m/s1s -9.8 m/s - -32 32 ft/sft/s

2s -19.6 m/s2s -19.6 m/s - -64 64 ft/sft/s

3s -29.4 m/s3s -29.4 m/s - -96 ft/s96 ft/s

4s4s - -3939..22 m/sm/s - -128 128 ft/sft/s

Displacement During free fall● Displacement = ½ a t2 for objects falling from

rest. a=accel due to gravity, -9.8m/s2.● After falling for 1 second, an object on Earth has

fallen -4.9m● After falling for 2 seconds, the object is -19.6m

below the drop position.● After falling for 3 seconds, the object is -44.1m

below the drop position.

**overhead model****overhead model**● as the ball goes as the ball goes upup gravity takes gravity takes 9.8 m/s9.8 m/s

offoff the velocity for the velocity for every secondevery second● as the ball goes as the ball goes downdown gravity gravity puts onputs on 9.8 9.8

m/sm/s for every second the ball falls for every second the ball falls

● Example problems p.68U Example problems p.68U ● Practice problem p.68U-V 3-7Practice problem p.68U-V 3-7

● reaction timer page 68Treaction timer page 68T

● Gravity is constant and therefore we Gravity is constant and therefore we can “sub” it into ALL of our formulas can “sub” it into ALL of our formulas

1.1. g = vg = v

t t

2. V2. Vf f = V= Vii + gt + gt

4. D = V4. D = Viit+ ½ gtt+ ½ gt22

5. V5. Vff2 2 = V= Vii

22 + 2 gd + 2 gd

*Class work**Class work*● The time the sky screamer ride at Astro world is free The time the sky screamer ride at Astro world is free

falling is 1.5 sec, (A) what is its velocity at the end of falling is 1.5 sec, (A) what is its velocity at the end of the time? (B) How far does it fall?the time? (B) How far does it fall?

GivenGiven FormulaFormula SolutionSolution

gg= = -- 9.8 m/s 9.8 m/s 2 2 VVff = V = Vii + + ggtt VVff= 0 + (= 0 + (--9.8)9.8)(1.5)(1.5)

VVii= 0 m/s d= V= 0 m/s d= Viit + ½ t + ½ ggtt2 2

t = 1t = 1..5 s d= 0 + ½ (5 s d= 0 + ½ (--9.8)9.8)(1.5)(1.5)22

(A): Vf= - 15 m/s

(B): d= -11 m

● A brick falls freely from a high scaffold (A) A brick falls freely from a high scaffold (A) what is the velocity after 4 seconds (B) how what is the velocity after 4 seconds (B) how far does the brick fall during these 4 seconds? far does the brick fall during these 4 seconds?

GivenGiven FormulaFormula SolutionSolution

g= - 9.8 m/sg= - 9.8 m/s2 2 VVff= V= Vii + gt V + gt Vff= 0 + = 0 + (-9.8)4 (-9.8)4

t= 4 sec d= Vt= 4 sec d= Viit + ½ gtt + ½ gt22 (A): V (A): Vff= -= -39.2m/s 39.2m/s

VVii= 0 m/s d= ½ (-9.8)= 0 m/s d= ½ (-9.8)(4)(4)22

( B): d= - ( B): d= - 78.4m78.4m

FormulasFormulasa=a= VVff – V – Vii VVii VVff aa tt

tt

VVff= V= Vii+ at+ at VVii VVf f a a tt

d= ½ (Vd= ½ (Vff + V + Vii)t)t VVii VVff dd tt

d= Vd= Viit + ½ att + ½ at22 VVii dd aa tt

VVff22= V= Vii

22 +2ad +2ad VVi i VVf f dd aa

g= Vg= Vff – V – Vii VVii VVf f gg tt

tt

VVff= V= Vii + gt + gt VVi i VVff gg tt

d= Vd= Viit + ½ gtt + ½ gt22 VVii dd g g tt

VVff22= V= Vii

22 + 2gd + 2gd VVii VVff dd gg

What’s going on here …..?

What’s going on here…..?