2.2 Acceleration

Biblical Reference

But flee there quickly, because I cannot do anything until you reach it.

Genesis 19:22

Flight of Lot

Bush 2011

Let’s Review

VELOCITY is the SLOPE of a distance, position, or displacement vs. time graph.

smslopeRun

Rise

t

xv /

seconds

meters

The RATE of CHANGE of VELOCITY

D = Change = FINAL - INITIAL

Dv = Final velocity – Initial velocity

2seconds

ondmeters/sec

s

m

t

va

Acceleration – The Definition

if vvv

Acceleration

2seconds

ondmeters/sec

s

m

t

va

The average acceleration of a particle is defined as the change in velocity Dvx divided by the time interval Dt during which that change occurred.

12

12

tt

vv

t

va xx

v

t

v1

v2

t1 t2

Dv

Dt

Acceleration is represented by the slope on a velocity-time graph

Average Acceleration

(s)

x

4

8

12

16

20

24

28

(m)

1 2 3 4t

5

1 2 3 4 5 t(s)

2

4

6

8

10

v

(m/s)sm

10v

s 5t

tv

a

s 5s

m10

2s

m 2

Displacement

25 m

Motion Diagrams

mss

mbhA 25)5)(10(

2

1

2

1

The slopes help us sketch the motion of other graphs!

Motion Diagrams

t

x

t

v

a

Displacement, velocity and acceleration graphs

The slope of a velocity-timegraph represents acceleration

tv

a

The slope of a displacement-timegraph represents velocity

tx

v

Motion Diagrams

t

x

t

v

t

a

Dt

Displacement, velocity and acceleration graphs

The area under an acceleration-timegraph represents change in velocity.

Dv

vta

The area under a velocity-time graph represents displacement.

Dx

xtv

Motion Diagrams

Comparing and Sketching graphs

It is often difficult to use one type of graph to draw another type of graph.

t (s)

x (m)

slope

= v List 2 adjectives to describe the SLOPE or VELOCITY

1. 2.

The slope is CONSTANTThe slope is POSITIVE

How could you translate what the SLOPE is doing on the graph ABOVE to the Y axis on the graph to the right?

t (s)

v (m/s)

Example

t (s)

x (m)

1st line• •

2nd line•

3rd line• •

The slope is constant

The slope is constant

The slope is “+”

The slope is “-”The slope is “0”

t (s)

v (m/s)

Example – Graph Matching

t (s)

v (m/s)

t (s)

a (m/s2)

t (s)

a (m/s2)

t (s)

a (m/s2)

What is the SLOPE(a) doing?The slope is increasing

The Big 5 (Kinematic Equations)Kinematics - Analyzing motion under the condition of constant acceleration.

Δd or Δx or Δy Change in Displacement

t Time

Vi or Vo Initial Velocity

Vf or V Final Velocity

a Acceleration

g Acceleration due to gravity (9.8 m/s2)

Equation # Equation Missing Variable

Equation 1Δx

Equation 2 Vf

Equation 3 a

Equation 4 t

Equation 5 Vi

atvv if

2

2

1attvx i

tx 2

vv fi

xavv if 222

2

2

1attvx f

An object moving with an initial velocity vi undergoes a constant acceleration a for a time t. Find the final velocity.

Δt

Δva Solution:

atv atvv o Eq 1

vi

time = 0 time = t

a?

atvv o

1-D Motion, Constant Acceleration

1 Eq atvv o

atvv o

atv

What are we calculating?

0 t

a

DV

1-D Motion, Constant Acceleration

An object moving with a velocity vo is passing position xo when it undergoes a constant acceleration a for a time t. Find the object’s displacement.

time = 0 time = t

xo?

a

vo

tx

vavg

x2

attv

2

o xtvavg

x0t2

vvo

x0t2

atvv oo

2at

tvx2

o Eq 2

atvv o 1 Eq

1-D Motion, Constant Acceleration

Solution:

What are we calculating?

2 Eq 2

attvx

2

o

tvo

2at2

0 t

vo

vt av

))(2

1(

2

1

2

1tatvtbhArea

1-D Motion, Constant Acceleration

atvv o Eq 1

2at

tvx2

o Eq 2

Solve Eq 1 for a and sub into Eq 2:

2t

t

v-vtvx

2o

o

Solve Eq 1 for t and sub into Eq 2:

2oo

o a

vv

2a

a

vvvx

t 2

vvx o

Eq 3

xa2vv 2o

2 Eq 4

t

vva o

a

vvt o

1-D Motion, Constant Acceleration

1. Read the problem carefully.

2. Sketch the problem.

3. Visualize the physical situation.

4. Identify the known and unknown quantities.

6. Use the variable that is not needed to pick the appropriate Big 5 equation.

7. Solve the equation for the variable needed.

9. Solve and check your answers.

5. Determine the variable that is not needed.

8. Plug quantities into equation.

Review – Problem Solving Skills

Example: A Cessna Aircraft goes from 0.00 m/s to 60.0 m/s in 13.0 seconds. Calculate the aircraft’s acceleration.

s

sm

t

va

13

0/604.62 m/s2

Given: vi = 0.00 m/s, vf = 60.0 m/s, Dt = 13.0 sec

Find: a (acceleration)

Example: The Cessna now decides to land and goes from 60.0 m/s to 0.00 m/s in 11.0 s. Calculate the Cessna’s acceleration.

deceleration

s

sm

t

va

11

/600- 5.45 m/s2

Given: vi = 60.0 m/s, vf = 0.00 m/s, Dt = 11.0 sec

Find: a (acceleration)

Example: You are designing an airport for small planes. One kind of plane that might use this airfield must reach a velocity before take off of at least 27.8 m/s and can accelerate at 2.00 m/s2.

a) If the runway is 150m long, can this airplane reach the required speed for take off?

b) If not, what minimum length must the runway have?

Given: vi = 0.00 m/s, vf = 27.8 m/s, a = 2.00 m/s2

Verify: Dx < 150. m

xavv if 222

a

vvx if

2

22

msm

smsmx 193

)/00.2(2

)/00.0()/8.27(2

22

The runway is too short.

Example: A roller coaster designer is designing a new roller coaster. A section of the coaster requires the cars that are traveling at 23.6 m/s must stop in 0.80 m.

The roller coaster designer needs to know if the amount of “g” force will be too great on the passengers. A “g” force of 16 g can be deadly.

1.00 g = 9.8 m/s2

16 g = 156.8 m/s2

Given: vi = 23.6 m/s, vf = 0.00 m/s, Dx = 0.80 m

Verify: a < 156.8 m/s2

xavv if 222

x

vva if

2

22

222

/1.348)80.0(2

)/6.23()/00.0(sm

m

smsma

The g-force is too great.

Given: a = 156 m/s2, vf = 0.00 m/s, Dx = 0.80 m

Find: vi

xavv if 222

15.8 m/s is a safe speed.

What is the maximum velocity that can safely stop within 0.80 m?

xavv fi 22

)80.0)(/156(2)/00.0( 22 msmsmvi

Franz Reuleaux