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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. Sign is very important!. Acceleration has both direction and magnitude - PowerPoint PPT Presentation
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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
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