Kinematics and velocity presentation that is very useful for your physics class

1. G O O D D AY !

2. W H AT I S O U R T O P I C ?

3. K I N E M AT I C S I N
O N E D I M E N S I O N
G E N E R A L P H Y S I C S 1
W E E K 2

4. T O P I C S
•Graphical Representations and Interpretations of Motion
•Motion Along a Straight Line
•Motion with Constant Acceleration
•Freely Falling Bodies

5. L E S S O N O B J E C T I V E S
At the end of the lesson, you should be able to:
K : : identify the kinematic variables (distance, time, velocity, and
acceleration) in each set of conditions of a particle in motion;
S : construct graphs given sets of values; : solve simple problems
involving uniform motion and uniformly accelerated motion
A : value the importance of graphs by understanding the pattern it
conveys especially in telling patterns and relationships.

8. • CONCEPTUAL COMPONENT - gravitational force
• MATHEMATICAL COMPONENT - F = ma

9. G R A P H I C A L
R E P R E S E N T A
T I O N S A N D
I N T E R P R E T A
T I O N S O F
M O T I O N

10. A . V E L O C I T Y V E R S U S T I M E

11. A . V E L O C I T Y V E R S U S T I M E
Area = length shown as time (s) x width shown as velocity (m/s) =
m or simply its displacement.

12. E X A M P L E 1 :
A particle is moving at 10 m/s, what is its displacement after 20
s?
Area = length x width
= 20 s x 10 m/s
= 200 m is the variable displacement

13. B . A C C E L E R AT I O N V E R S U S T I M E
Area = length shown as time(s) x width shown as acceleration =
m/s or its velocity.

14. E X A M P L E 2 :
A particle is accelerating at 3.0 m/s2, what is its velocity after 20
seconds?
Area = 20 s x 3.0 m/s2
= 60 m/s is the variable velocity

15. C . P O S I T I O N V E R S U S T I M E
Slope = y component (m) ÷ x component (s) = m/s, which is the
unit of measure for velocity (m/s)

16. E X A M P L E 3 :
A particle has moved from 0 to 30 m after 10 seconds. What is
its velocity?
m = y/x
= 30 m/ 10 s
= 3 m/s which is the particle’s velocity and represents
graphically the slope of the line

17. D . V E L O C I T Y V E R S U S T I M E
Slope (m) = y component ÷ x component = velocity (m/s) ÷ time (s)
= m/s2 .
This will give us the physical quantity called acceleration in (m/s2 ).

18. E X A M P L E 4 :
A particle is moving from 0 m/s to 20 m/s. After 10 seconds what
is its acceleration? Using the formula to find the slope of the line,
m = y component ÷ x component.
m = (20 - 0) m/s ÷ (10 - 0) s
= 2 m/s2, which is a graphical representation of the linear
slope and its corresponding physical quantity the acceleration.

19. M O T I O N
A L O N G A
S T R A I G H T
L I N E

20. D I S TA N C E A N D D I S P L A C E M E N T
Distance is a scalar quantity that refers to “how much ground
an object has covered” during its motion. Displacement is a
vector quantity that refers to “how far out of place an object
is”; it is the object's overall change in position.

21. • A particle moving along the x-axis has a coordinate x.
• The change in the particle’s coordinate is ∆x = x2 − x1.
• ∆x is not the product of ∆ and x. “∆” means change in the
quantity. ∆x is the change in the quantity of x.
• ∆x can be positive or negative.

22. AV E R A G E S P E E D
Speed is a scalar quantity.
• Has the same unit as velocity.
• Defined as total distance/total time:
The speed has no direction and is always expressed as a positive
number.

23. AV E R A G E V E L O C I T Y
The average x-velocity is change in x divided by the time interval:

24. N E G AT I V E V E L O C I T Y
The average velocity is negative during a time interval if the
particle moves in the negative x-direction for that time
interval.

25. I N S TA N TA N E O U S V E L O C I T Y
Instantaneous velocity is the velocity at a specific instant or time or a specific
point in the path. It indicates what is happening at every point of time.
It is the limit of the average velocity as the interval approaches 0. It is equal to
the instantaneous rate of change of position with time.

26. • Your neighbor’s dog is crouched 25 m to your east. At time, the dog
begins to run due east toward a rat that is 50 m to your east. During
the first 2.0 s of the attack, the dog’s coordinate 𝑥 varies with time
according to the following equation:

27. 1. Find the dog’s displacement between = 1.5 s and = 2.0 s.
2. Find its average velocity during that interval.
3. Derive an expression for the dog’s instantaneous velocity as
a function of time and use it to find 𝑥 at = 1.5 s and = 2.0 s.
The change in the dog’s displacement on the interval t = 1.5 s to
t = 2.0s is ∆x = x2 − x1 = 45.00 m − 36.25 m = 8.75 m.

28. 2. The average velocity on the interval t1 = 1.5 s to t2 = 2.0 s is
3. The instantaneous x velocity of the dog is

29. 3. The instantaneous x velocity of the dog is
Constant Rule
f(x) = c
f’(x)= 0
y=2 so dy =0
dx
Power Rule
f(x) = xn where n is any real number
f’(x) = n xn-1
(xn)’= n xn-1

30. 𝑑
𝑑𝑡
𝑥 =
𝑑
𝑑𝑡
25𝑚 +
𝑑
𝑑𝑡
5
𝑚
𝑠2
𝑡2
𝑑
𝑑𝑡
𝑥 = 0 + 2 5
𝑚
𝑠2
𝑡2−1
𝑑
𝑑𝑡
𝑥 = 0 + 10
𝑚
𝑠2
𝑡1
Vx = 10
𝑚
𝑠2 𝑡

31. At 1.5 s
At 2.0 s

32. AV E R A G E V S . I N S TA N TA N E O U S
A C C E L E R AT I O N
Acceleration is the rate of change of velocity with time.

33. 2. The average velocity on the interval t1 = 1.5 s to t2 = 2.0 s is
3. The instantaneous x velocity of the dog is
First, we derive an expression for the vx from the given equation
above:

34. 3. The instantaneous x velocity of the dog is
First, we derive an expression for the vx from the given equation
above:

35. E X A M P L E
Suppose the x velocity of the pedicab in Figure at any time t is given
by the equation:

36. 1. Find the change in x velocity of the pedicab in the time
interval = 1.0 s to t = 2.0 s.
2. Find the average x acceleration in this interval.
3. Derive the expression for x acceleration as a function of
time and use it to find ax at t = 1.0 s and t = 2.0 s.

37. S O L U T I O N
1. Before we can find the change in velocity at the time interval, we
must find the velocities at each time t from the given equation:
The change in x velocity ∆vx between t = 1.0 s and t = 2.0 s is
- = 1.5 m/s

38. 2. The average x acceleration aave−x at the given time interval is
1.5 m/s2
m/s2 , since s is x 1

39. 3. The instantaneous x acceleration is
First, we derive an expression for the ax which is

40. At t = 1.0 s
At t = 2.0 s

41. E X A M P L E
Suppose the x-velocity of the car in Fig. 2.11 at any time t is given by
the equation:

42. (a) Find the change in x-velocity of the car in the time interval t1=
1.0s and t2 = 3.0s
(b) Find the average x-acceleration in this time interval.
(c) Find the instantaneous x-acceleration at time t1= 1.0s by taking
Δt to be first 0.1 s, then 0.01 s, then 0.001 s.
(d) Derive an expression for the instantaneous x-acceleration as a
function of time, and use it to find at t = 1.0s and t = 3.0s.

43. A. Find the change in x-velocity of the car in the time interval
t1= 1.0s and t2 = 3.0s
The change in x-velocity between t1= 1.0s and t2 = 3.0s is

44. B. Find the average x-acceleration in this time interval t1=
1.0s and t2 = 3.0s is

45. C. When , we have t2 = 1.1s . Proceeding as
before, we find

46. D. Derive an expression for the instantaneous x-acceleration
as a function of time, and use it to find at t = 1.0s and t =
3.0s.

48. M O T I O N W I T H C O N S TA N T
A C C E L E R AT I O N
An accelerating object will change its velocity by the same
amount each second. This is referred to as a constant
acceleration since the velocity is changing by a constant
amount each second.

50. K I N E M AT I C F O R M U L A S

51. C O N S TA N T - A C C E L E R AT I O N C A L C U L AT I O N
A pedicab heading east through a small town accelerates at a
constant 4 after leaving the city limits. At time t = 0, it is
5.0 m east of the city limit signpost, moving east at 15 m/s.
(a) Find the position and velocity at t = 2.0 s.
(b) Where is it when velocity is 25 m/s?

54. T H A N K Y O U !