Acceleration and Free Fall
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Transcript of Acceleration and Free Fall
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Acceleration and Free Fall
Chapter 2.2 and 2.3
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What is acceleration?
Acceleration measures the rate of change in velocity.
Average acceleration = change in velocity/ time required for change
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Units for acceleration
2
/
s
m
s
sm
t
vaavg
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Sign is very important!
Acceleration has both direction and magnitude
A negative value for acceleration does not always mean an object is decelerating!!
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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
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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
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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
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The Kinematic EquationsYou are going to loooooove these!
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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.
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Displacement with constant acceleration
Δx = displacementVi = initial velocity
Vf = final velocity
Δt = time interval
tvvx fi )(2
1
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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!!!
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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
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Velocity with constant uniform acceleration
tavv if
Vf = final velocityVi = initial velocitya = accelerationΔt = time interval
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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
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Solve
Plug in values and solve for Vf
Vf= 19 m/s
)0.5)(0.3(3.42
ss
m
s
mv f
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Displacement with constant uniform acceleration
2)(2
1tatvx i
Δx = displacementVi = initial velocitya = accelerationΔt = time interval
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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
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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
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Final Velocity after any displacement
xavv if 222
Vf = final velocityVi = initial velocitya = accelerationΔx = displacement
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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
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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
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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!)
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Section 2-3 Falling Objects
Free Fall: Neglecting air resistance, all objects fall with the same constant acceleration
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Acceleration due to gravity
281.9s
mg
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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
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What we see because of air resistance…
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Object falling from rest
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Path of a projectile
At top of pathv= 0 m/sa = -9.81 m/s2
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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
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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?
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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
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Solve the problem
xavv if 22
)25)(81.9(202
2
ms
m
s
mv f
s
mv f 1.22
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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
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Solve the Problem
tavv if
ta
vv if
281.9
01.22
smsm
t
st 25.2