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Linear motion
The motion of an object along
a straight path is called
a linear motion.
A train running on a straight railway track
Vehicles moving on a straight highway
Sprinters running on a straight athletic track
Motion of a bowling ball
Motion diagrams
Frame-1
Motion diagrams
Frame-2
Motion diagrams
Frame-3
Motion diagrams
Frame-4
Motion diagram of a car moving towards a tree
Motion diagrams
Motion diagrams
A racing car running on a track
Images that are equally spaced indicate an
object moving with constant spe...
Motion diagrams
A cyclist starting a race
An increasing distance between the images
shows that the object is speeding up.
Motion diagrams
A car stopping for a red light
A decreasing distance between the images
shows that the object is slowing d...
Particle model
1
Motion diagram of a racing car moving with constant speed
2 3 4
Same motion diagram using the particle mo...
Position
The location of an object
at a particular instant is
called its position.
Position
Your school
Your position
EW
(reference point)
(direction)
4 km
(distance)
Coordinate system
The school defines the origin of
the coordinate system
1-1 0 2 3 4
Your position is at
+4 km
(km)
Case-I: Position is zero
The car is at origin
x = 0 m
-20 -10 0 10 20 30 40 50 60 (m)
x
Case-II: Position is positive
The car is at positive side of origin
x = 50 m
-20 -10 0 10 20 30 40 50 60
x
(m)
x
Case-III: Position is negative
x = −40 m
The car is at negative side of origin
-50 -40 -30 -20 -10 0 10 20 30
x
(m)
x
The displacement of an object is the
change in the position of an object
in a particular direction.
∆x = xf − xi
Displacem...
Case-I: Displacement is positive
∆x = 50 − (−20)
∆x = 70 m
If object is moving towards
right, the displacement is
positive...
Case-II: Displacement is negative
-20 -10 0 10 20 30 40 50 60
xf xiB A
∆x = −10 − (40)
∆x = −50 m
If object is moving towa...
Case-III: Displacement is zero
∆x = 40 − (40)
∆x = 0 m
If object come to its initial
position, the displacement is
zero be...
Distance
The actual path covered by an
object is called a distance.
d = actual path covered
Distance
∆xAB = xB − xA
= 50 − 10
= 40 = 40 m
= −10 − 50
= −60 = 60 m
d = ∆xAB + ∆xBC
= 40 + 60
= 100 m
-20 -10 0 10 20 30...
The rate of change of position of an object in a given
direction is called its velocity.
Velocity
Velocity =
displacement
...
Case-I: Velocity is positive
Velocity is positive because change in position
is positive.
0 10 20 30 40 50 60 70 80 (m)
x
...
Case-II: Velocity is negative
0 10 20 30 40 50 60 70 80 (m)
x
0 s1 s2 s3 s4 s
∆x = −15 m ∆x = −15 m ∆x = −15 m ∆x = −15 m
...
Case-III: Velocity is zero
0 10 20 30 40 50 60 70 80 (m)
x
0 s, 1 s, 2 s, 3 s, 4s
∆x = 0 m
Velocity is zero because change...
Speed
The rate at which an object covers a distance is
called a speed.
Speed =
distance
time
• Scalar quantity (only magni...
0 10 20 30 40 50 60 70 80 (m)
x
0 s
d = 15 m
Case-I: Speed is positive
1 s 2 s 3 s 4 s
speed =
15
1
= 15 m/s
If an object ...
Case-II: Speed is zero
0 10 20 30 40 50 60 70 80 (m)
x
0 s, 1 s, 2 s, 3 s, 4s
d = 0 m
If an object is at rest the distance...
Average velocity
0 10 20 30 40 50 60 70 80 (m)
x
A
vAB =
10 − 0
1 − 0
=
10
1
= 10 m/s
vBC =
30 − 10
2 − 1
=
20
1
= 20 m/s
...
Average velocity
For an object moving with variable velocity,
average velocity is defined as the ratio of its
total displa...
Average velocity
vavg =
10 + 20 + 40 − 30
1 + 1 + 1 + 1
vavg =
40
4
= 10 m/s
0 10 20 30 40 50 60 70 80 (m)
x
A vAB
2 s
4 s...
Average velocity
If x1 and x2 are the positions of an object at
times t1 and t2, then the average velocity from
time t1 to...
Average velocity
vavg =
40 − 0
4 − 0
=
40
4
vavg = 10 m/s
0 10 20 30 40 50 60 70 80 (m)
x
A vAB
2 s
4 s 3 s
B C D
E
0 s 1 ...
Average speed
SAB =
10 − 0
1 − 0
=
10
1
= 10 m/s
SBC =
30 − 10
2 − 1
=
20
1
= 20 m/s
SCD =
70 − 30
3 − 2
=
40
1
= 40 m/s
S...
For an object moving with variable speed, the
average speed is the total distance travelled by
the object divided by the t...
Average speed
Savg =
10 + 20 + 40 + 30
1 + 1 + 1 + 1
Savg =
100
4
= 25 m/s
0 10 20 30 40 50 60 70 80 (m)
xSABA
0 s
B
1 s 2...
Instantaneous velocity
The velocity of an object at a particular instant
of time or at a particular point of its path is
c...
Instantaneous velocity
x
t1 , x1
A
t2 , x2
B
∆x
∆t
x
t1 , x1
A
t2 , x2
B
∆x
∆t
C
C
vC
vC
Instantaneous velocity
x
t1 , x1
A
t2 , x2
B
∆x
∆t
C
xC
dx
dt
vC
vC
Instantaneous velocity
Instantaneous velocity is equal to the limiting
value of the average velocity of the object in a
sm...
Instantaneous speed
The magnitude of instantaneous
velocity is called instantaneous speed.
0 10 20 30 40 50 60 70 80 (m)
x...
Acceleration
The rate of change of velocity of an object with
time is called its acceleration.
acceleration =
change in ve...
Case-I: Velocity is constant
acceleration = 0 m/s2
Acceleration is zero because change in velocity
is zero.
v = 20 m/s v =...
acceleration = 5 m/s2
Case-II: Velocity is increasing towards right
Acceleration is positive because change in velocity
is...
Case-III: Velocity is decreasing towards right
acceleration = −4 m/s2
Acceleration is negative because change in velocity
...
acceleration = −3 m/s2
Case-IV: Velocity is increasing towards left
Acceleration is negative because change in velocity
is...
acceleration = 6 m/s2
Case-V: Velocity is increasing towards left
Acceleration is positive because change in velocity
is p...
Important note
• If the signs of the velocity and acceleration of
an object are the same, the speed of the
object increase...
Average acceleration
aAB =
15 − 20
2 − 0
=
−5
2
= −2.5 m/s2
aDE =
−10 − 30
8 − 6
=
−40
2
= −20 m/s2
aCD =
30 − (−5)
6 − 4
...
Average acceleration
For an object moving with variable velocity, the
average acceleration is defined as the ratio of
the ...
Average acceleration
aavg =
−5 − 20 + 35 − 40
2 + 2 + 2 + 2
aavg =
−30
8
= −3.75 m/s2
vD = 30 m/svB = 15 m/s vC = −5 m/svA...
If v1 and v2 are the velocities of an object at
times t1 and t2, then the average acceleration
from time t1 to t2 is given...
Average acceleration
aavg =
−10 − 20
8 − 0
=
−30
8
aavg = −3.75 m/s2
vD = 30 m/svB = 15 m/s vC = −5 m/svA = 20 m/s
0 s 8 s...
Instantaneous acceleration
The acceleration of an object at a particular
instant of time or at a particular point of its
p...
Instantaneous acceleration
t
t1 , v1
A
t2 , v2
B
∆v
∆t
t
t1 , v1
A
t2 , v2
B
∆v
∆t
C
C aC
aC
Instantaneous acceleration
t
t1 , v1
A
t2 , v2
B
∆v
∆t
C
tC
dv
dt
aC
aC
Instantaneous acceleration
Instantaneous acceleration is equal to the limiting
value of the average acceleration of the ob...
Uniformly accelerated motion
The motion in which the velocity of
an object changes with uniform rate
or constant rate is c...
Uniformly accelerated motion
A car slowing down after a red signal
A car speeding up after a green signal
Kinematical equations of motion
xi
T = 0
xf
T = t
x
vi vf
1. Velocity after a certain time:
acceleration =
change in veloc...
Kinematical equations of motion
2. Displacement in a certain time:
By definition,
average velocity =
displacement
time
vav...
Kinematical equations of motion
3. Velocity after certain displacement:
We know that,
vf = vi + a t
… … . . (1)vf − vi = a...
Equations of motion by calculus method
1. First equation of motion:
By definition,
a =
dv
dt
dv = a dt … … . . (1)
When ti...
Equations of motion by calculus method
2. Second equation of motion:
By definition,
v =
dx
dt
dx = v dt … … . . (1)
When t...
Equations of motion by calculus method
3. Second equation of motion:
By definition,
a =
dv
dt
v dv = a dx … … . . (1)
When...
Graphical
representation
of motion
Position-time graph
Time (s) Position (m)
0 30
5 55
10 40
15 0
20 -35
25 -55
5 10 15 20 25 30
t0
-40
-20
20
40
60
-60
x
Average velocity from position-time graph
5 10 15 20 25 30
t0
-40
-20
20
40
60
-60
x
vavg(AB) =
−55 − 55
25 − 5
=
−110
20
...
Instantaneous velocity from position-time graph
5 10 15 20 25 30
t0
-40
-20
20
40
60
-60
∆x
∆t
A
B
(5,5)
(25,-55)θ
Slope o...
Analyzing nature of motion through position–time graph
t
x
A B
slope zero
v = 0 x
0 v = 0
The particle is at rest on posit...
t
x
θ
θ
θ
Analyzing nature of motion through position–time graph
A
B
C
slope positive
v > 0
x
0
The particle is at negativ...
Analyzing nature of motion through position–time graph
t
x
θ
θ
θ
slope negative
v < 0
x
0
The particle is at positive side...
Analyzing nature of motion through position–time graph
t
x
θ1
θ2
θ3
x
0
The particle is at negative side of origin and
mov...
Analyzing nature of motion through position–time graph
t
x
θ1
θ2
θ3
A
B
C
x
0
The particle is at negative side of origin a...
Analyzing nature of motion through position–time graph
t
x
θ1
θ2
θ3
x
0
The particle is at positive side of origin and
mov...
Analyzing nature of motion through position–time graph
t
x
θ1
θ2
θ3
x
0
The particle is at positive side of origin and
mov...
Velocity-time graph
Time (s) velocity (m/s)
0 -40
5 0
10 20
15 10
20 -20
25 -40
5 10 15 20 25 30
t0
-40
-20
20
40
60
-60
v
5 10 15 20 25 30
t0
-40
-20
20
40
60
-60
Average acceleration from velocity-time graph
aavg(AB) =
−40 − 0
25 − 5
=
−40
20
...
5 10 15 20 25 30
t0
-40
-20
20
40
60
-60
Instantaneous acceleration from velocity-time graph
aB =
−40 − 7
25 − 10
=
−47
15...
Analyzing nature of motion through velocity–time graph
t
v
A B
slope zero
a = 0 x
v
The particle is moving with
constant v...
Analyzing nature of motion through velocity–time graph
t
v
θ
θ
θ
A
B
C
slope positive
a > 0
x
The particle is moving towar...
Analyzing nature of motion through velocity–time graph
t
v
θ
θ
θ
slope negative
a < 0
A
B
C
x
The particle is moving towar...
t
v
θ1
θ2
θ3
slope positive
a > 0
A
B
C
Analyzing nature of motion through velocity–time graph
x
The particle is moving to...
Analyzing nature of motion through velocity–time graph
t
v
θ1
θ2
θ3
A
B
C
slope positive
a > 0
x
For portion AB:
For porti...
Analyzing nature of motion through position–time graph
t
v
θ1
θ2
θ3
A
B
C
slope negative
a < 0
x
For portion AB:
For porti...
t
v
θ1
θ2
θ3
A
B
C
Analyzing nature of motion through position–time graph
x
For portion AB:
For portion BC:
v3v2v1
x
v4v5v...
Acceleration–time graph
Time (s)
acceleration
(m/s 𝟐
)
0 20
5 50
10 30
15 0
20 -30
25 -40 5 10 15 20 25 30
t0
-40
-20
20
4...
Analyzing nature of motion through acceleration–time graph
t
a
A B
slope zero
∆a = 0 x
The particle is moving with
constan...
Analyzing nature of motion through acceleration–time graph
t
a
θ
θ
θ
A
B
C
slope positive
∆a > 0
x
The acceleration of par...
Analyzing nature of motion through acceleration–time graph
t
a
θ
θ
θ
slope negative
∆a < 0
A
B
C
x
The acceleration of par...
Graphical
integration in
motion analysis
t
a
a
dt
area of strip = a dt
ti tf
dv = a dt
From graph,
By definition,
Net change in velocity from time ti to tf ,
vi
vf...
t
v
v
dt
area of strip = v dt
ti tf
dx = v dt
From graph,
By definition,
Net change in position from time ti to tf ,
xi
xf...
Thank
you
Linear motion of a particle
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Linear motion of a particle

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This PPT covers linear motion of an object in a very systematic and lucid manner. I hope this PPT will be helpful for instructor's as well as students.

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Linear motion of a particle

  1. 1. Linear motion The motion of an object along a straight path is called a linear motion.
  2. 2. A train running on a straight railway track
  3. 3. Vehicles moving on a straight highway
  4. 4. Sprinters running on a straight athletic track
  5. 5. Motion of a bowling ball
  6. 6. Motion diagrams Frame-1
  7. 7. Motion diagrams Frame-2
  8. 8. Motion diagrams Frame-3
  9. 9. Motion diagrams Frame-4
  10. 10. Motion diagram of a car moving towards a tree Motion diagrams
  11. 11. Motion diagrams A racing car running on a track Images that are equally spaced indicate an object moving with constant speed.
  12. 12. Motion diagrams A cyclist starting a race An increasing distance between the images shows that the object is speeding up.
  13. 13. Motion diagrams A car stopping for a red light A decreasing distance between the images shows that the object is slowing down.
  14. 14. Particle model 1 Motion diagram of a racing car moving with constant speed 2 3 4 Same motion diagram using the particle model
  15. 15. Position The location of an object at a particular instant is called its position.
  16. 16. Position Your school Your position EW (reference point) (direction) 4 km (distance)
  17. 17. Coordinate system The school defines the origin of the coordinate system 1-1 0 2 3 4 Your position is at +4 km (km)
  18. 18. Case-I: Position is zero The car is at origin x = 0 m -20 -10 0 10 20 30 40 50 60 (m) x
  19. 19. Case-II: Position is positive The car is at positive side of origin x = 50 m -20 -10 0 10 20 30 40 50 60 x (m) x
  20. 20. Case-III: Position is negative x = −40 m The car is at negative side of origin -50 -40 -30 -20 -10 0 10 20 30 x (m) x
  21. 21. The displacement of an object is the change in the position of an object in a particular direction. ∆x = xf − xi Displacement
  22. 22. Case-I: Displacement is positive ∆x = 50 − (−20) ∆x = 70 m If object is moving towards right, the displacement is positive because xf > xi -20 -10 0 10 20 30 40 50 60 ∆x xfxiA B (m) x
  23. 23. Case-II: Displacement is negative -20 -10 0 10 20 30 40 50 60 xf xiB A ∆x = −10 − (40) ∆x = −50 m If object is moving towards left, the displacement is negative because xf < xi (m) x ∆x
  24. 24. Case-III: Displacement is zero ∆x = 40 − (40) ∆x = 0 m If object come to its initial position, the displacement is zero because xf = xi -20 -10 0 10 20 30 40 50 60 xf xi A B (m) x
  25. 25. Distance The actual path covered by an object is called a distance. d = actual path covered
  26. 26. Distance ∆xAB = xB − xA = 50 − 10 = 40 = 40 m = −10 − 50 = −60 = 60 m d = ∆xAB + ∆xBC = 40 + 60 = 100 m -20 -10 0 10 20 30 40 50 60 A BC (m) x ∆xBC = xC − xB
  27. 27. The rate of change of position of an object in a given direction is called its velocity. Velocity Velocity = displacement time • Vector quantity (magnitude & direction). • SI unit is metre/second (m/s). • Dimensional formula is [L 1 T −1 ].
  28. 28. Case-I: Velocity is positive Velocity is positive because change in position is positive. 0 10 20 30 40 50 60 70 80 (m) x 0 s ∆x = 20 m ∆x = 20 m ∆x = 20 m ∆x = 20 m 1 s 2 s 3 s 4 s v = 20 m/s velocity = 20 1 = 20 m/s
  29. 29. Case-II: Velocity is negative 0 10 20 30 40 50 60 70 80 (m) x 0 s1 s2 s3 s4 s ∆x = −15 m ∆x = −15 m ∆x = −15 m ∆x = −15 m Velocity is negative because change in position is negative. velocity = −15 1 = −15 m/s v = −15 m/s
  30. 30. Case-III: Velocity is zero 0 10 20 30 40 50 60 70 80 (m) x 0 s, 1 s, 2 s, 3 s, 4s ∆x = 0 m Velocity is zero because change in position is zero. velocity = 0 1 = 0 m/s
  31. 31. Speed The rate at which an object covers a distance is called a speed. Speed = distance time • Scalar quantity (only magnitude). • SI unit is metre/second (m/s). • Dimensional formula is [L 1 T −1 ].
  32. 32. 0 10 20 30 40 50 60 70 80 (m) x 0 s d = 15 m Case-I: Speed is positive 1 s 2 s 3 s 4 s speed = 15 1 = 15 m/s If an object moves it cover some distance and hence the speed is positive. d = 15 m d = 15 m d = 15 m
  33. 33. Case-II: Speed is zero 0 10 20 30 40 50 60 70 80 (m) x 0 s, 1 s, 2 s, 3 s, 4s d = 0 m If an object is at rest the distance covered is zero and hence the speed is zero. speed = 0 1 = 0 m/s
  34. 34. Average velocity 0 10 20 30 40 50 60 70 80 (m) x A vAB = 10 − 0 1 − 0 = 10 1 = 10 m/s vBC = 30 − 10 2 − 1 = 20 1 = 20 m/s vCD = 70 − 30 3 − 2 = 40 1 = 40 m/s vDE = 40 − 70 4 − 3 = −30 1 = −30 m/s vAB 2 s 4 s 3 s B C D E 0 s 1 s D vBC vCD vDE
  35. 35. Average velocity For an object moving with variable velocity, average velocity is defined as the ratio of its total displacement to the total time interval in which that displacement occurs. Average velocity = total displacement total time
  36. 36. Average velocity vavg = 10 + 20 + 40 − 30 1 + 1 + 1 + 1 vavg = 40 4 = 10 m/s 0 10 20 30 40 50 60 70 80 (m) x A vAB 2 s 4 s 3 s B C D E 0 s 1 s D vBC vCD vDE
  37. 37. Average velocity If x1 and x2 are the positions of an object at times t1 and t2, then the average velocity from time t1 to t2 is given by vavg = x2 − x1 t2 − t1 = ∆x ∆t
  38. 38. Average velocity vavg = 40 − 0 4 − 0 = 40 4 vavg = 10 m/s 0 10 20 30 40 50 60 70 80 (m) x A vAB 2 s 4 s 3 s B C D E 0 s 1 s D vBC vCD vDE
  39. 39. Average speed SAB = 10 − 0 1 − 0 = 10 1 = 10 m/s SBC = 30 − 10 2 − 1 = 20 1 = 20 m/s SCD = 70 − 30 3 − 2 = 40 1 = 40 m/s SDE = 40 − 70 4 − 3 = −30 1 = 30 m/s 0 10 20 30 40 50 60 70 80 (m) xSABA 0 s B 1 s 2 s C D 3 s D 4 s E SBC SCD SDE
  40. 40. For an object moving with variable speed, the average speed is the total distance travelled by the object divided by the total time taken to cover that distance. Average speed = total distance total time Average speed
  41. 41. Average speed Savg = 10 + 20 + 40 + 30 1 + 1 + 1 + 1 Savg = 100 4 = 25 m/s 0 10 20 30 40 50 60 70 80 (m) xSABA 0 s B 1 s 2 s C D 3 s D 4 s E SBC SCD SDE
  42. 42. Instantaneous velocity The velocity of an object at a particular instant of time or at a particular point of its path is called its instantaneous velocity. 0 10 20 30 40 50 60 70 80 (m) x 0 s 1 s vA = 15 m/s vB = 20 m/s vC = 10 m/s vD = −5 m/sA B 2 s C D 3 s
  43. 43. Instantaneous velocity x t1 , x1 A t2 , x2 B ∆x ∆t x t1 , x1 A t2 , x2 B ∆x ∆t C C vC vC
  44. 44. Instantaneous velocity x t1 , x1 A t2 , x2 B ∆x ∆t C xC dx dt vC vC
  45. 45. Instantaneous velocity Instantaneous velocity is equal to the limiting value of the average velocity of the object in a small time interval taken around that instant, when time interval approaches zero. v = lim ∆t→0 x2 − x1 t2 − t1 = lim ∆t→0 ∆x ∆t = dx dt
  46. 46. Instantaneous speed The magnitude of instantaneous velocity is called instantaneous speed. 0 10 20 30 40 50 60 70 80 (m) x 0 s SA = 15 m/s SB = 20 m/s SC = 10 m/s SD = 5 m/s 1 s 2 s 3 s A B C D
  47. 47. Acceleration The rate of change of velocity of an object with time is called its acceleration. acceleration = change in velocity time • Vector quantity (magnitude & direction). • SI unit is metre/second/second (m/s2 ). • Dimensional formula is [L 1 T −2 ].
  48. 48. Case-I: Velocity is constant acceleration = 0 m/s2 Acceleration is zero because change in velocity is zero. v = 20 m/s v = 20 m/s v = 20 m/s v = 20 m/s 0 s 1 s 2 s 3 s x ∆v = 0 m/s ∆v = 0 m/s ∆v = 0 m/s a = 0 m/s2 a = 0 m/s2 a = 0 m/s2
  49. 49. acceleration = 5 m/s2 Case-II: Velocity is increasing towards right Acceleration is positive because change in velocity is positive. a = 5 m/s2 vA = 5 m/s vB = 10 m/s vC = 15 m/s vD = 20 m/s 0 s 1 s 2 s 3 s ∆v = 5 m/s x ∆v = 5 m/s ∆v = 5 m/s a = 5 m/s2 a = 5 m/s2
  50. 50. Case-III: Velocity is decreasing towards right acceleration = −4 m/s2 Acceleration is negative because change in velocity is negative. vA = 20 m/s vB = 16 m/s vC = 12 m/s vD = 8 m/s 0 s 1 s 2 s 3 s x ∆v = −4 m/s a = −4 m/s2 ∆v = −4 m/s a = −4 m/s2 ∆v = −4 m/s a = − 4 m/s2
  51. 51. acceleration = −3 m/s2 Case-IV: Velocity is increasing towards left Acceleration is negative because change in velocity is negative. vD = −12 m/s vC = −9 m/s vB = −6 m/s vA = −3 m/s 0 s3 s 2 s 1 s x ∆v = −3 m/s a = −3 m/s2 ∆v = −3 m/s a = −3 m/s2 ∆v = −3 m/s a = −3 m/s2
  52. 52. acceleration = 6 m/s2 Case-V: Velocity is increasing towards left Acceleration is positive because change in velocity is positive. vA = −24 m/svB = −18 m/svC = −12 m/svD = −6 m/s 3 s 0 s1 s2 s x ∆v = 6 m/s a = 6 m/s2 ∆v = 6 m/s a = 6 m/s2 ∆v = 6 m/s a = 6 m/s2
  53. 53. Important note • If the signs of the velocity and acceleration of an object are the same, the speed of the object increases. • If the signs are opposite, the speed decreases.
  54. 54. Average acceleration aAB = 15 − 20 2 − 0 = −5 2 = −2.5 m/s2 aDE = −10 − 30 8 − 6 = −40 2 = −20 m/s2 aCD = 30 − (−5) 6 − 4 = 35 2 = 17.5 m/s2 aBC = −5 − 15 4 − 2 = −20 2 = −10 m/s2 vD = 30 m/svB = 15 m/s vC = −5 m/svA = 20 m/s 0 s 8 s2 s x ∆vAB = −5 m/s ∆vBC = −20 m/s ∆vCD = 35 m/s 4 s 6 s vE = −10 m/s ∆vDE = −40 m/s A B C D E
  55. 55. Average acceleration For an object moving with variable velocity, the average acceleration is defined as the ratio of the total change in velocity of the object to the total time interval taken. Average acceleration = total change in velocity total time
  56. 56. Average acceleration aavg = −5 − 20 + 35 − 40 2 + 2 + 2 + 2 aavg = −30 8 = −3.75 m/s2 vD = 30 m/svB = 15 m/s vC = −5 m/svA = 20 m/s 0 s 8 s2 s x ∆vAB = −5 m/s ∆vBC = −20 m/s ∆vCD = 35 m/s 4 s 6 s vE = −10 m/s ∆vDE = −40 m/s A B C D E
  57. 57. If v1 and v2 are the velocities of an object at times t1 and t2, then the average acceleration from time t1 to t2 is given by aavg = v2 − v1 t2 − t1 = ∆v ∆t Average acceleration
  58. 58. Average acceleration aavg = −10 − 20 8 − 0 = −30 8 aavg = −3.75 m/s2 vD = 30 m/svB = 15 m/s vC = −5 m/svA = 20 m/s 0 s 8 s2 s x ∆vAB = −5 m/s ∆vBC = −20 m/s ∆vCD = 35 m/s 4 s 6 s vE = −10 m/s ∆vDE = −40 m/s A B C D E
  59. 59. Instantaneous acceleration The acceleration of an object at a particular instant of time or at a particular point of its path is called its instantaneous acceleration. 0 10 20 30 40 50 60 70 80 (m) x 0 s 1 s 2 s 3 s aA = 25 m/s2 aB = 15 m/s2 aC = 10 m/s2 aD = −5 m/s2 A B C D
  60. 60. Instantaneous acceleration t t1 , v1 A t2 , v2 B ∆v ∆t t t1 , v1 A t2 , v2 B ∆v ∆t C C aC aC
  61. 61. Instantaneous acceleration t t1 , v1 A t2 , v2 B ∆v ∆t C tC dv dt aC aC
  62. 62. Instantaneous acceleration Instantaneous acceleration is equal to the limiting value of the average acceleration of the object in a small time interval taken around that instant, when time interval approaches zero. a = lim ∆t→0 v2 − v1 t2 − t1 = lim ∆t→0 ∆v ∆t = dv dt = d dt dx dt = d2 x dt2
  63. 63. Uniformly accelerated motion The motion in which the velocity of an object changes with uniform rate or constant rate is called uniformly accelerated motion.
  64. 64. Uniformly accelerated motion A car slowing down after a red signal A car speeding up after a green signal
  65. 65. Kinematical equations of motion xi T = 0 xf T = t x vi vf 1. Velocity after a certain time: acceleration = change in velocity time By definition, a = vf − vi t − 0 vf − vi = a t vf = vi + a t This is the first kinematical equation.
  66. 66. Kinematical equations of motion 2. Displacement in a certain time: By definition, average velocity = displacement time vavg = xf − xi t − 0 But, vavg = vi + vf 2 ∴ xf − xi t = vi + vf 2 xf − xi t = 2vi 2 + a t 2 xf − xi t = vi + 1 2 a t xf − xi = vit + 1 2 a t2 This is the second kinematical equation. xf − xi t = vi + vi + at 2 = 2vi + a t 2
  67. 67. Kinematical equations of motion 3. Velocity after certain displacement: We know that, vf = vi + a t … … . . (1)vf − vi = a t Also, vf + vi 2 = xf − xi t vf + vi = xf − xi 2 t … … . . (2) multiplying equations (1) & (2), we get (vf − vi)(vf + vi) = a t (xf − xi) 2 t L. H. S = vf 2 − vi 2 R. H. S = 2 a (xf − xi) This is the third kinematical equation. vf 2 − vi 2 = 2 a (xf − xi) vf 2 = vi 2 + 2 a (xf − xi)
  68. 68. Equations of motion by calculus method 1. First equation of motion: By definition, a = dv dt dv = a dt … … . . (1) When time = 0, velocity = vi When time = t, velocity = vf vi vf dv = 0 t a dt Integrating equation (1) within the above limits of time & velocity, we get v vi vf = a t 0 t = a 0 t dt vf − vi = a (t − 0) vf − vi = a t vf = vi + a t
  69. 69. Equations of motion by calculus method 2. Second equation of motion: By definition, v = dx dt dx = v dt … … . . (1) When time = 0, position = xi When time = t, position = xf Integrating equation (1) within the above limits of time & position, we get = 0 t (vi + a t) dt xf − xi = vi t − 0 + 1 2 a (t2 − 02) xi xf dx = 0 t vi dt + 0 t a t dt x xi xf = vi t 0 t + a t2 2 0 t xf − xi = vit + 1 2 a t2 xi xf dx = 0 t v dt
  70. 70. Equations of motion by calculus method 3. Second equation of motion: By definition, a = dv dt v dv = a dx … … . . (1) When time = 0, velocity = vi, position = xi Integrating equation (1) within the above limits of velocity & position, we get v2 2 vi vf = a x xi xf = dv dx × dx dt = dv dx × v When time = t, velocity = vf, position = xf v2 vi vf = 2a x xi xf vf 2 − vi 2 = 2 a (xf − xi) vf 2 = vi 2 + 2 a (xf − xi) vi vf v dv = xi xf a dx
  71. 71. Graphical representation of motion
  72. 72. Position-time graph Time (s) Position (m) 0 30 5 55 10 40 15 0 20 -35 25 -55 5 10 15 20 25 30 t0 -40 -20 20 40 60 -60 x
  73. 73. Average velocity from position-time graph 5 10 15 20 25 30 t0 -40 -20 20 40 60 -60 x vavg(AB) = −55 − 55 25 − 5 = −110 20 vavg(AB) = −5.5 m/s Slope of line AB gives average velocity between points A & B. slope of line AB = tan θ tan θ = ∆x ∆t = x2 − x1 t2 − t1 = vavg ∆x ∆t A B (5,55) (25,-55)θ
  74. 74. Instantaneous velocity from position-time graph 5 10 15 20 25 30 t0 -40 -20 20 40 60 -60 ∆x ∆t A B (5,5) (25,-55)θ Slope of tangent at point B gives instantaneous velocity at point B. vB = −55 − 5 25 − 5 = −60 20 vB = −3 m/s slope of tangent = tan θ tan θ = ∆x ∆t = x2 − x1 t2 − t1 = v
  75. 75. Analyzing nature of motion through position–time graph t x A B slope zero v = 0 x 0 v = 0 The particle is at rest on positive side of origin.
  76. 76. t x θ θ θ Analyzing nature of motion through position–time graph A B C slope positive v > 0 x 0 The particle is at negative side of origin and moving with constant velocity towards origin. For portion AB: For portion BC: x 0 The particle is at origin and moving with constant velocity towards right side of origin. v v
  77. 77. Analyzing nature of motion through position–time graph t x θ θ θ slope negative v < 0 x 0 The particle is at positive side of origin and moving with constant velocity towards origin. For portion AB: For portion BC: x 0 The particle is at origin and moving with constant velocity towards left side of origin. A B C v v
  78. 78. Analyzing nature of motion through position–time graph t x θ1 θ2 θ3 x 0 The particle is at negative side of origin and moving with increasing velocity towards origin. For portion AB: For portion BC: x 0 The particle is at origin and moving with increasing velocity towards right side of origin. v v slope positive v > 0 A B C a a
  79. 79. Analyzing nature of motion through position–time graph t x θ1 θ2 θ3 A B C x 0 The particle is at negative side of origin and moving with decreasing velocity towards origin. For portion AB: For portion BC: x 0 The particle is at origin and moving with decreasing velocity towards right side of origin. v vslope positive v > 0 a a
  80. 80. Analyzing nature of motion through position–time graph t x θ1 θ2 θ3 x 0 The particle is at positive side of origin and moving with increasing velocity towards origin. For portion AB: For portion BC: x 0 The particle is at origin and moving with increasing velocity towards left side of origin. v v A B C slope negative v < 0 a a
  81. 81. Analyzing nature of motion through position–time graph t x θ1 θ2 θ3 x 0 The particle is at positive side of origin and moving with decreasing velocity towards origin. For portion AB: For portion BC: x 0 The particle is at origin and moving with decreasing velocity towards left side of origin. v v A B C a a slope negative v < 0
  82. 82. Velocity-time graph Time (s) velocity (m/s) 0 -40 5 0 10 20 15 10 20 -20 25 -40 5 10 15 20 25 30 t0 -40 -20 20 40 60 -60 v
  83. 83. 5 10 15 20 25 30 t0 -40 -20 20 40 60 -60 Average acceleration from velocity-time graph aavg(AB) = −40 − 0 25 − 5 = −40 20 aavg(AB) = −2 m/s2 Slope of line AB gives average acceleration between points A & B. slope of line AB = tan θ tan θ = ∆v ∆t = v2 − v1 t2 − t1 = aavg v ∆v ∆t B (25,-40) A (5,0) θ
  84. 84. 5 10 15 20 25 30 t0 -40 -20 20 40 60 -60 Instantaneous acceleration from velocity-time graph aB = −40 − 7 25 − 10 = −47 15 aB = −3.14 m/s2 Slope of line AB gives instantaneous acceleration between points A & B. slope of tangent = tan θ tan θ = ∆v ∆t = v2 − v1 t2 − t1 = a v ∆v ∆t B (25,-40) A (5,7) θ
  85. 85. Analyzing nature of motion through velocity–time graph t v A B slope zero a = 0 x v The particle is moving with constant velocity towards right. v a = 0
  86. 86. Analyzing nature of motion through velocity–time graph t v θ θ θ A B C slope positive a > 0 x The particle is moving towards left and its velocity is uniformly decreasing. For portion AB: For portion BC: v1v2v3 aaa The particle is moving towards right and its velocity is uniformly increasing. x v6v5v4 aaa
  87. 87. Analyzing nature of motion through velocity–time graph t v θ θ θ slope negative a < 0 A B C x The particle is moving towards right and its velocity is uniformly decreasing. For portion AB: For portion BC: v3v2v1 aaa The particle is moving towards left and its velocity is uniformly increasing. x v6 v5 v4 aaa
  88. 88. t v θ1 θ2 θ3 slope positive a > 0 A B C Analyzing nature of motion through velocity–time graph x The particle is moving towards left and its velocity is decreasing slowly at non-uniform rate. For portion AB: For portion BC: v1v2v3 x v6v5v4 a1a2a3 a4 a5 a6 The particle is moving towards right and its velocity is increasing rapidly at non-uniform rate.
  89. 89. Analyzing nature of motion through velocity–time graph t v θ1 θ2 θ3 A B C slope positive a > 0 x For portion AB: For portion BC: v1v2v3 x v6v5v4 a1a2a3 a6a5a4 The particle is moving towards left and its velocity is decreasing rapidly at non-uniform rate. The particle is moving towards right and its velocity is increasing slowly at non-uniform rate.
  90. 90. Analyzing nature of motion through position–time graph t v θ1 θ2 θ3 A B C slope negative a < 0 x For portion AB: For portion BC: v3v2v1 x v4v5v6 a3a2a1 a6 a5 a4 The particle is moving towards right and its velocity is decreasing slowly at non-uniform rate. The particle is moving towards left and its velocity is increasing rapidly at non-uniform rate.
  91. 91. t v θ1 θ2 θ3 A B C Analyzing nature of motion through position–time graph x For portion AB: For portion BC: v3v2v1 x v4v5v6 a3a2a1 a6 a5 a4 The particle is moving towards right and its velocity is decreasing rapidly at non-uniform rate. The particle is moving towards left and its velocity is increasing slowly at non-uniform rate.
  92. 92. Acceleration–time graph Time (s) acceleration (m/s 𝟐 ) 0 20 5 50 10 30 15 0 20 -30 25 -40 5 10 15 20 25 30 t0 -40 -20 20 40 60 -60 a
  93. 93. Analyzing nature of motion through acceleration–time graph t a A B slope zero ∆a = 0 x The particle is moving with constant acceleration. a a
  94. 94. Analyzing nature of motion through acceleration–time graph t a θ θ θ A B C slope positive ∆a > 0 x The acceleration of particle is uniformly decreasing towards left. For portion AB: For portion BC: x a1a2a3 The acceleration of particle is uniformly increasing towards right. a4 a5 a6
  95. 95. Analyzing nature of motion through acceleration–time graph t a θ θ θ slope negative ∆a < 0 A B C x The acceleration of particle is uniformly decreasing towards right. For portion AB: For portion BC: x a3a2a1 The acceleration of particle is uniformly increasing towards left. a6 a5 a4
  96. 96. Graphical integration in motion analysis
  97. 97. t a a dt area of strip = a dt ti tf dv = a dt From graph, By definition, Net change in velocity from time ti to tf , vi vf dv = ti tf a dt vf − vi = area between acceleration curve and time axis, from ti to tf
  98. 98. t v v dt area of strip = v dt ti tf dx = v dt From graph, By definition, Net change in position from time ti to tf , xi xf dx = ti tf v dt xf − xi = area between velocity curve and time axis, from ti to tf
  99. 99. Thank you
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This PPT covers linear motion of an object in a very systematic and lucid manner. I hope this PPT will be helpful for instructor's as well as students.

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