Acceleration and Free Fall

Chapter 2.2 and 2.3

What is acceleration?

Acceleration measures the rate of change in velocity.

Average acceleration = change in velocity/ time required for change

Units for acceleration

2

/

s

m

s

sm

t

vaavg

Sign is very important!

Acceleration has both direction and magnitude

A negative value for acceleration does not always mean an object is decelerating!!

Speeding up, moving to the left

Slowing down, moving to the rightSpeeding up, moving to the right

2-4 Acceleration

Increasing speed and deceleration (decreasing speed) should not be confused with the directions of velocity and acceleration:

Slowing down, moving to the left

Fill in the Chart

Initial Velocity Acceleration Motion+ + Speeding up, moving

right/up

- - Speeding up, moving left/down

+ - Slowing Down moving right/up

- + Slowing Down, moving left/down

- or + 0 Constant Velocity

0 - or + Speeding up from rest

0 0 Remaining at rest

Graph of Velocity vs Time

Question: What does the slope of this graph give you?

Rise = ΔvRun Δt

Answer: ACCELERATION

Vf – VAVG = Δv

tf – ti = Δt

The Kinematic EquationsYou are going to loooooove these!

Motion with constant acceleration

Kinematic EquationsThe relationships between displacement,

velocity and constant acceleration are expressed by equations that apply to any object moving with constant acceleration.

Displacement with constant acceleration

Δx = displacementVi = initial velocity

Vf = final velocity

Δt = time interval

tvvx fi )(2

1

Example: #1 p.53 in book

A car accelerates uniformly from rest to a speed of 23.7 km/h in 6.5 s. Find the distance the car travels during this time.Δx = displacement= distance= ?Vi = initial velocity = rest = 0 km/h

Vf = final velocity = 23.7 km/hΔt = time interval = 6.5 s

Look at final velocity…convert to m/s!!!

Problem Solving

Final velocity

conversion

Plug in values and solve for Δx

2371000

1

1

3600658. .

km

hx

m

kmx

h

s

m

s

mss

m

s

mx 21)5.6)(58.60(2

1

Velocity with constant uniform acceleration

tavv if

Vf = final velocityVi = initial velocitya = accelerationΔt = time interval

Example: #2 p.55

An automobile with an initial speed of 4.30 m/s accelerates uniformly at the rate of 3.0 m/s2. Find the final speed after 5.0 seconds.

Vf = final velocity=?Vi = initial velocity = 4.3 m/sa = acceleration= 3.0 m/s^2Δt = time interval= 5.0 s

Solve

Plug in values and solve for Vf

Vf= 19 m/s

)0.5)(0.3(3.42

ss

m

s

mv f

Displacement with constant uniform acceleration

2)(2

1tatvx i

Δx = displacementVi = initial velocitya = accelerationΔt = time interval

Example: #2 p.55

An automobile with an initial speed of 4.30 m/s accelerates uniformly at the rate of 3.0 m/s2. Find the displacement after 5.0 seconds.

Δx = displacement=??

Vi = initial velocity= 4.30 m/s

a = acceleration= 3.0 m/s^2

Δt = time interval= 5.0 s

Solve!

Plug in values and solve for displacement

mss

ms

s

mx 59)0.5)(0.3(

2

1)0.5)(3.4( 2

2

Final Velocity after any displacement

xavv if 222

Vf = final velocityVi = initial velocitya = accelerationΔx = displacement

Example: p.58 #3

A car accelerates uniformly in a straight line from rest at the rate of 2.3 m/s^2. What is the speed of the car after it has traveled 55 m?

Vf = final velocity=??Vi = initial velocity= rest= 0 m/sa = acceleration= 2.3 m/s^2Δx = displacement= 55 m

Solve

s

mv

s

mv

s

mm

s

m

s

mv

ff

f

16253

253)55)(3.2(2)0(

2

22

2

2

222

Rearranging

Your problems won’t always be so straightforward…make sure to rearrange your equations to solve for the unknown before plugging in your numbers (with units!)

Section 2-3 Falling Objects

Free Fall: Neglecting air resistance, all objects fall with the same constant acceleration

Acceleration due to gravity

281.9s

mg

Free Fall Acceleration

• However, acceleration is a vector.

• Gravity acts toward the earth (down)

• Therefore, the acceleration of objects in free fall near the surface of the earth is

281.9s

mga

What we see because of air resistance…

Object falling from rest

Path of a projectile

At top of pathv= 0 m/sa = -9.81 m/s2

Free Fall Acceleration

At the highest point of an arc, an object has velocity = 0 m/s, acceleration is still -9.81 m/s2

An object thrown into the air is a freely falling body with

s

mvi 0

Free Fall Problem p.64 #2

A flowerpot falls from a windowsill 25.0 m above the sidewalkA. How fast is the flowerpot moving when it

strikes the ground?

B. How much time does a paserby on the sidewalk below have to move out of the way before the flowerpot hits the ground?

Part. A.

What are we looking for: Vf

What do we know? Displacement: -25 m Acceleration: -9.81 m/s2

Vi=0 m/s

What equation should we use??xavv if 222

Solve the problem

xavv if 22

)25)(81.9(202

2

ms

m

s

mv f

s

mv f 1.22

Part b.

How much time before the flowerpot hits the ground? What do we know? Displacement= -25.0 m Acceleration = -9.81

m/s2

V initial= 0 V final = -22.1 m/s What are we looking

for: Time! Which equation should

we use??

tavv if

Solve the Problem

tavv if

ta

vv if

281.9

01.22

smsm

t

st 25.2