1. Accelerated Motion
Chapter 3

2. Chapter Objectives Describe accelerated motion
Use graphs and equations to solve problems involving moving objects
Describe the motion of objects in free fall (separate presentation - not on this test)

3. Section 3.1 Acceleration Define acceleration
Relate velocity and acceleration to the motion of an object
Create velocity-time graphs

4. Uniform Motion Nonuniform Motion
Moving at a constant velocity
If you close your eyes, you feel as though you are not moving at all
Moving while changing velocity
Can be changing the rate or the direction
You feel like you are being pushed or pulled

5. Changing Velocity
Consider the following motion (particle model) diagram
Constant Velocity
Not moving
Increasing Velocity
Decreasing Velocity

6. Changing velocity You can indicate change in velocity by
the motion diagram spacing
the magnitude (length) of the velocity vectors.
If the object speeds up, each subsequent velocity vector is longer.
If the object slows down, each vector is shorter than the previous one.

7. Velocity-Time Graphs
Distance being covered is longer, thus the runner is speeding up.
Distance being covered is longer, thus the runner is speeding up.

8. Velocity-Time Graphs Time (s) Velocity (m/s) 1 5 2 10 3 15 4 20 25
Slope???
Time (s)
Velocity (m/s) 1 5 2 10 3 15 4 20 25
Area???

9. Change in Position → can’t get position
Velocity-Time Graphs
Analyze the units
Slope = rise over run
m = ∆y / ∆x
Slope = m/s/ s = m/s^2
m/s^2 is the unit for acceleration
Area = ½ b h
b = s * m/s = m
m is the unit for displacement
The slope of a velocity-time graph is the ACCELERATION and the area is DISPLACEMENT.
Change in Position → can’t get position

10. Area =
Displacement
Slope =
Acceleration

11. Velocity – Time Graphs How Fast something is moving at a given time?
Average Acceleration
Use the information on the x & y axis to plug into the equation
a = ∆v / t
Slope
Instantaneous Acceleration
Find the slope of the line (straight line)
Find the slope of the tangent (curve)
Displacement
Find the area under the curve
You do not know the initial or final position of the runner, just the displacement.

12. Velocity-Time Graphs Describe the motion of each sprinter. A
Constant velocity
Zero Acceleration
Positive displacement B
Constant Acceleration
Starts from Rest
Positive displacement
Describe the motion of each sprinter.

13. Velocity-Time Graphs D Constant Acceleration Positive Acceleration
Comes to a Stop
Zero displacement E
Constant Velocity
Zero Acceleration
Negative displacement C
Constant Acceleration
Negative Acceleration
Comes to a Stop
Positive displacement

14. Sample Question
What is the average acceleration of the car shown on the graph below?
A. 0.5 m/s2
B m/s2
C. 2 m/s2
D. -2 m/s2

15. Acceleration The rate at which an object’s velocity changes
Variable: a
Units: m/s2
It is the change in velocity which measures the change in position. Thus, it is measuring a change of a change, hence why the square time unit.
When the velocity of an object changes at a constant rate, it has constant acceleration

16. Motion Diagrams & Acceleration
In order for a motion diagram to display a full picture of an object’s movement, it should contain information about acceleration by including average acceleration vectors.
The vectors are average acceleration vectors because motion diagrams display the object at equal time INTERVALS (intervals always mean average)
Average acceleration vectors are found by subtracting two consecutive velocity vectors.

17. Average Acceleration Vectors
You will have: Δv = vf - vi = vf + (-vi).
Then divide by the time interval, Δt. The time interval, Δt, is 1 s. This vector, (vf - vi)/1 s, shown in violet, is the average acceleration during that time interval.

18. Average Acceleration Vectors
vi = velocity at the beginning of a chosen time interval
vf = velocity at the end of a chosen time interval.
∆v = change in velocity
* Acceleration is equal to the change in velocity over the time interval
** Since the time interval is 1s, the acceleration is equal to the change in velocity
***Anything divided by 1 is equal to itself…

19. Average vs. Instantaneous Acceleration
Average Acceleration
Change in velocity during some measurable time interval divided by the time interval
Found by plugging into the equation
Instantaneous Acceleration
Change in velocity at an instant of time
Found by calculating the slope of a velocity-time graph at that instant
𝐚= ∆𝒗 𝒕

20. Velocity & Acceleration
How would you describe the sprinter’s velocity and acceleration as shown on the graph?

21. Velocity & Acceleration
Sprinter’s velocity starts at zero
Velocity increases rapidly for the first four seconds until reaching about 10 m/s
Velocity remains almost constant

22. Average vs. Instantaneous Acceleration
What is the acceleration for the first four seconds?
Refers to average acceleration because there is a time interval
Solve using the equation a = ∆v /t
vi = 0 m/s; vf = 11 m/s; t = 4s
a = (11m/s – 0 m/s)/ 4s
a = 2.75 m/s2

23. Average vs. Instantaneous Acceleration
What is the acceleration at 5s?
Refers to instantaneous acceleration because it is looking for acceleration at an instant
Need to find the slope of the line to solve for acceleration
Slope is zero; thus instantaneous acceleration is zero at the instant of 5s.

24. Instantaneous Acceleration
Solve for the acceleration at 1.0 s
Draw a tangent to the curve at t = 1s
The slope of the tangent is equal to the instantaneous acceleration at 1s.
a = rise / run

25. Instantaneous Acceleration
The slope of the line at 1.0 s is equal to the acceleration at that instant .

26. Positive & Negative Acceleration
These four motion diagrams represent the four different possible ways to move along a straight line with constant acceleration.

27. Object is moving in the positive direction
Displacement is positive
Thus, velocity is positive
Object is getting faster
Acceleration is positive

28. Object is moving in the positive direction
Displacement is positive
Thus, velocity is positive
Object is getting slower
Acceleration is negative

29. Object is moving in the negative direction
Displacement is negative
Thus, velocity is negative
Object is getting faster
Acceleration is negative

30. Object is moving in the negative direction
Displacement is negative
Thus, velocity is negative
Object is getting slower
Acceleration is positive

31. Positive & Negative Acceleration
When the velocity vector and acceleration vector point in the SAME direction, the object is INCREASING SPEED
When the velocity vector and acceleration vector point in the OPPOSITE direction, the object is DECREASING SPEED

32. + UP - Down Displacement Velocity Acceleration Speeding UP Or
Displacement & Velocity always have the same sign
Displacement
Velocity
Acceleration
Speeding UP Or
Slowing Down + UP -
Down
Up = same
Down = Different

33. Sample Question
How can the instantaneous acceleration of an object with varying acceleration be calculated?
A. by calculating the slope of the tangent on a distance versus time graph
B. by calculating the area under the graph on a distance versus time graph
C. by calculating the area under the graph on a velocity versus time graph
D. by calculating the slope of the tangent on a velocity versus time graph

34. Practice v-t graph A B C D E

35. Segment t
(s)
vi (m/s) vf
(m/s) ∆v
avg. a
(m/s2)
ins. A Xi
(m) Xf ∆X A 0  B C D E
**Can not assume position on graph. Velocity time graphs can only be used to figure out displacement. You must be given an initial position.

36. 3.2 Motion with Constant Acceleration
Interpret position-time graphs for motion with constant acceleration
Determine mathematical relationships among position, velocity, acceleration, and time
Apply graphical and mathematical relationships to solve problems related to constant acceleration.

37. Constant acceleration: x-t Graphs
Velocity is constantly increasing, which means more displacement.
Slope must be getting steeper.
Results in a curve that is parabolic.

38. Constant acceleration: x-t Graphs
x (m)
t (s)
Concave UP = +a
t (s)
Concave UP = + a
x (m)
x (m)
t (s)
Concave Down = -a
Concave Down = -a
x (m)
t (s)

39. Kinematics Equations
Three equations that relate position, velocity, acceleration, and time.
First two are derived from a v-t graph and the third is a substitution.
Total of five different variables.
Δx (displacement), vi (initial velocity), vf (final velocity), a (acceleration), and t (time).
Must know any three in order to solve for the other two.

40. First Kinematics Equation
Remember that the slope of a v-t graph is the average acceleration.
Rearranging the equation, gives us the first kinematics equation.
Replace tf – ti with t
vf = vi + at

41. Second Kinematics Equation
We remember that area of a v-t graph equals displacement
Break into two known shapes (rectangle & triangle).
Total Area = Area of rectangle + area of triangle
Δx = vit + ½ (vf –vi)t
vf – vi = at
(substitute)
Δx = vit + ½ at2

42. Third Kinematics Equation
First equation substituted into the second to cancel out the time variable.
vf = vi + at t = (vf – vi) / a
Δx = vit + ½ at2
Δx = vi((vf – vi)/a) + ½ a ((vf – vi)/a) 2
Δx = vivf – vi2 + ½ a (vf2 – 2 vivf + vi2 )/a2
2a Δx = 2 vivf - 2 vi2 + vf2 – 2 vivf + vi2
2a Δx = - vi2 + vf2 (rearrange)
Simplify
Multiply by 2a to get rid of fraction
Combine like terms
vf2 = vi2 + 2a Δx

43. No displacement (position)
Kinematics
Equation
Use when:
NO ∆x
No displacement (position)
No vf
No final velocity
No t
No time
vf2 = vi2 + 2a Δx
Δx = vit + ½ at2
vf = vi + at
Must have 3 of the 5 variables to start. (∆x, vi, vf, a, t)
Must always know acceleration.
If acceleration is not given, then must solve for it first.

44. Sample Problem Given Unknown Formula Substitution Answer
An automobile starts at rest and speeds up at 3.5 m/s2 after the traffic light turns green. How far will it have gone when it is traveling at 25 m/s?
vi = 0 m/s
vf = 25 m/s
a = 3.5 m/s2
∆x = ? m
vf2 = vi2 + 2a ∆x
(25m/s)2 = (0 m/s)2 + 2 (3.5 m/s2)(∆x )
∆x = 89.3 m
Given
Unknown
Formula
Substitution
Answer

45. Sample Problem
You are driving a car, traveling at constant velocity of 25 m/s, when you see a child suddenly run onto the road. It takes you 0.45 s to react and apply the brakes. As a result, the car slows with a steady acceleration of 8.5 m/s2 and comes to a stop. What is the total distance that the car moves before it stops?
Reaction
Stopping
t =0.45 s
v = 25 m/s
vi = 25 m/s
vf = 0 m/s
a = -8.5 m/s2
∆xreaction = ? m
∆xstopping = ? m
𝑣= ∆𝑥 𝑡
25 𝑚 𝑠 = ∆𝑥 0.45𝑠
(0 m/s)2 = (25 m/s)2 + 2(-8.5m/s2)(∆x)
∆x = m
∆xstopping = m
vf2 = vi2 + 2a Δx
xtotal = m