Uniformly Accelerated Motion (horizontally and Vertically)

Uniformly Accelerated Motion (horizontally and Vertically)

1. Acceleration and Free Fall

2. What is acceleration?
Acceleration measures the rate of
change in velocity.
Average acceleration = change in
velocity/ time required for change

3. Units for acceleration
2
/
s
m
s
sm
t
v
a
avg





4. Sign is very important!
Acceleration has both direction and
magnitude
A negative value for acceleration does
not always mean an object is
decelerating!!

5. 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

6. Fill in the Chart
Initial VelocityAccelerationMotion
+ +
Speeding up, moving
right/up
- -
Speeding up, moving
left/down
+ -
Slowing Down
moving right/up
- +
Slowing Down,
moving left/down
0
Constant Velocity
0 - or +
Speeding up from
rest
0 0
Remaining at rest

7. Fill in the Chart
Initial VelocityAccelerationMotion
+ +
Speeding up, moving
right/up
- -
Speeding up, moving
left/down
+ -
Slowing Down moving
right/up
- +
Slowing Down,
moving left/down
0
Constant Velocity
0 - or +
Speeding up from rest
0 0
Remaining at rest

8. Graph of Velocity vs Time
Question: What does the slope of this graph give you?
Rise = Δv
Run Δt
Answer: ACCELERATION
V
f
– V
AVG
= Δv
t
f
– t
i
= Δt

9. The Kinematic Equations
You are going to loooooove these! 

10. Motion with constant acceleration
Kinematic Equations
The relationships between displacement,
velocity and constant acceleration are
expressed by equations that apply to any
object moving with constant
acceleration.

11. Displacement with constant acceleration
Δx = displacement
V
i = initial velocity
V
f = final velocity
Δt = time interval
tvvx
fi  )(
2
1

12. Example: #1
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= ?
V
i = initial velocity = rest = 0 km/h
V
f
= final velocity = 23.7 km/h
Δt = time interval = 6.5 s
Look at final velocity…convert to m/s!!!

13. Problem Solving
Final velocity
conversion
Plug in values and
solve for Δx

237
1000
1
1
3600
658. .
km
h
x
m
km
x
h
s
m
s

ms
s
m
s
m
x 21)5.6)(58.60(
2
1


14. Velocity with constant uniform acceleration
tavv
if 
V
f = final velocity
V
i
= initial velocity
a = acceleration
Δt = time interval

15. Example:
An automobile with an initial speed of
4.30 m/s accelerates uniformly at the
rate of 3.0 m/s
2
. Find the final speed
after 5.0 seconds.
V
f = final velocity=?
V
i = initial velocity = 4.3 m/s
a = acceleration= 3.0 m/s^2
Δt = time interval= 5.0 s

16. Solve
Plug in values and
solve for Vf
V
f
= 19 m/s
)0.5)(0.3(3.4
2
s
s
m
s
m
v
f


17. Displacement with constant uniform acceleration
2
)(
2
1
tatvx
i

Δx = displacement
V
i
= initial velocity
a = acceleration
Δt = time interval

18. Example:
An automobile with an initial speed of 4.30 m/s
accelerates uniformly at the rate of 3.0 m/s
2
.
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

19. Solve!
Plug in values and
solve for
displacement
ms
s
m
s
s
m
x 59)0.5)(0.3(
2
1
)0.5)(3.4(
2
2


20. Final Velocity after any displacement
xavv
if  2
22
V
f
= final velocity
V
i
= initial velocity
a = acceleration
Δx = displacement

21. Example:
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/s
a = acceleration= 2.3 m/s^2
Δx = displacement= 55 m

22. Solve
s
m
v
s
m
v
s
m
m
s
m
s
m
v
ff
f
16253
253)55)(3.2(2)0(
2
2
2
2
2
2
22



23. 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!)

24. Section 2-3 Falling Objects
Free Fall: Neglecting air resistance, all
objects fall with the same constant
acceleration

25. Acceleration due to gravity
2
81.9
s
m
g

26. 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
2
81.9
s
m
ga 

27. What we see because of air
resistance…

28. Object falling from rest

29. Path of a projectile
At top of path
v= 0 m/s
a = -9.81 m/s
2

30. Free Fall Acceleration
At the highest point of an arc, an object has
velocity = 0 m/s, acceleration is still -9.81
m/s
2
An object thrown into the air is a freely
falling body with
s
m
v
i
0

31. Free Fall Problem
A flowerpot falls from a windowsill 25.0
m above the sidewalk
A. 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?

32. Part. A.
What are we looking for: V
f
What do we know?
Displacement: -25 m
Acceleration: -9.81 m/s
2
V
i
=0 m/s
What equation should we use??
xavv
if
 2
22

33. Solve the problem 
xavv
if  2
2
)25)(81.9(20
2
2
m
s
m
s
m
v
f







s
m
v
f
1.22

34. Part b.
How much time before
the flowerpot hits the
ground?
What do we know?
Displacement= -25.0 m
Acceleration = -9.81
m/s
2
V initial= 0
V final = -22.1 m/s
What are we looking
for: Time!
Which equation should
we use??
tavv
if


35. Solve the Problem 
tavv
if 
t
a
vv
if


2
81.9
01.22
s
m
s
m
t



st25.2

36. QUIZ

37. QUIZ

38. QUIZ

39. QUIZ

40. QUIZ

41. QUIZ

42. QUIZ

43. QUIZ

44. QUIZ

45. QUIZ

46. Solve the Problem 

47. Solve the Problem 

48. ANSWER

49. ANSWER

50. ANSWER