18. Difference between instantaneous speed and
velocities
instantaneous speed measures how fast a particle is
moving at a particular time while instantaneous
velocity measures how fast a particle is moving in a
specific direction at a particular time
26. EXERCISE 1
• Atrain changes its position (x) as a function of time (t)
as follows: x = 10 +3t2, with x m.
• (a) find the displacement of the train between t1 = 1.0 s
and t2 = 3.0 s.
• (b) find the average velocity during the same interval.
• (c) find the instantaneous velocity at the time t1 = 3.0 s
• (d) find the acceleration of the train.
27.
28. EXERCISE 2
• The position of a projectile travelling in 2D space is
given by x(t)= (2t2 + 4t + 3)m and y(t) = (3t2 – 5t + 2)m.
calculate the magnitude and direction of the projectile’s
displacement and the average velocity between time
interval t =2 s and t = 5 s.
36. Kinematics for the VERTICAL Direction
All 3 kinematics can be used to analyze one dimensional
motion in either the X direction OR the y direction.
v=vo +at →vy =voy +gt
v2
=v2
+2a(x−x )→v2
=v2
+2g(y−y )
ox o y oy o
2 2
1 1 2
2
o oy
x=xo +voxt+ gt
at →y= y +v t+
37. “g” or ag – The Acceleration due to gravity
The acceleration due to gravity is a special constant that exists in a
VACUUM, meaning without air resistance. If an object is in FREE
FALL, gravity will CHANGE an objects velocity by 9.8 m/s every
second.
2
g = ag = −9.8 m / s
The acceleration due to gravity:
•ALWAYS ACTS DOWNWARD
•IS ALWAYS CONSTANT near the
surface of Earth
42. Examples
What do I
know?
What do I
want?
voy= 0 m/s y = ?
g = -9.8 m/s2
yo=0 m
t = 5.78 s
A stone is dropped at rest from the top of a cliff. It is
observed to hit the ground 5.78 s later. How high is the
cliff?
Which variable is NOT given and
NOT asked for?
Final Velocity!
2
o oy
y = y + v t + 1 gt2
y =
y = (0)(5.78) − 4.9(5.78)2
-163.7 m
H =163.7m
43. Examples
A pitcher throws a fastball with a velocity of 43.5 m/s. It is determined that
during the windup and delivery the ball covers a displacement of 2.5
meters. This is from the point behind the body when the ball is at rest to
the point of release. Calculate the acceleration during his throwing
motion.
What do I
know?
What do I
want?
vo= 0 m/s a = ?
x = 2.5 m
v = 43.5 m/s
Which variable is NOT given and
NOT asked for?
TIME
v2
= v2
+ 2a(x − x )
o o
43.52
= 02
+ 2a(2.5− 0)
a = 378.5 m/s/s
44. Examples
What do I
know?
What do I
want?
vo= 0 m/s t = ?
x = 35 m
a = 2.00 m/s/s
How long does it take a car at rest to cross a 35.0 m intersection after
the light turns green, if the acceleration of the car is a constant
2.00 m/s/s?
Which variable is NOT given and
NOT asked for?
Final Velocity
35 = 0 + (0) + 1
2 (2)t2
t = 5.92 s
2
2
1 at
x = xo + voxt +
45. Examples
What do I
know?
What do I
want?
vo= 12.5 m/s a = ?
v = 25 m/s
t = 6s
Acar accelerates from 12.5 m/s to 25 m/s in 6.0 seconds.
What was the acceleration?
Which variable is NOT given and
NOT asked for?
DISPLACEMENT
o
v = v + at
a =
25 =12.5+ a(6)
2.08 m/s/s
46. Kinematics and Calculus
Let’s take the “derivative” of
kinematic #2 assuming the
object started at x = 0.
dx
dt dt
ox
ox
=
v =
at )
2
1
d(v t +
2
v = vo + at
a =
dv
=
d(v0 + at)
= a
dt dt
2
x = v t + 1 at2
47. Projectile Motion
Projectile: Any object which is projected by some means
and continues to move due to its own inertia (mass).
Aprojectile is an object moving in two dimensions under
the influence of Earth’s gravity; its path is a parabola.
Projectile motion can be understood by analyzing the
horizontal and vertical motions separately.
48. Projectiles move in TWO dimensions
Since a projectile moves
in 2-dimensions, it
therefore has 2
components just like
a resultant vector.
• Horizontal and
Vertical
49. Horizontally Launched Projectiles
To analyze a projectile in 2 dimensions we need 2
equations. One for the “x” direction and one for the “y”
direction.And for this we use kinematic #2.
2
ox
x = v t + 1 at2
x = voxt
Remember, the velocity
is CONSTANT
horizontally, so that
means the acceleration
is ZERO!
y = 1
2 gt2
Remember that since the
projectile is launched
horizontally, the INITIAL
VERTICAL VELOCITY is
equal to ZERO.
50. Horizontally Launched Projectiles
What do I
know?
What I want to
know?
vox=100 m/s t = ?
y = 500 m x = ?
voy= 0 m/s
g = -9.8 m/s/s
2
102.04 = t → t = 10.1 seconds
Example: A plane traveling with a
horizontal velocity of 100 m/s is
500 m above the ground.At
some point the pilot decides to
drop some supplies to designated
target below. (a) How long is the
drop in the air? (b) How far
away from point where it was
launched will it land?
y = 1
2 gt2
→ −500 = 1
2 (−9.8)t2
x = voxt = (100)(10.1) = 1010 m
51. Vertically Launched Projectiles
• For the horizontal
motion x-direction
vx = vox
x = voxt
• For the vertical motion
y-direction
vy = voy − gt
2
oy
y = v t −
1
gt2
52. Vertically Launched Projectiles
You will still use kinematic #2, but YOU MUST use
COMPONENTS in the equation.
vo
vox
voy
x=voxt 2
oy
y=v t+ 1 gt2
ox o
v = v cos
voy = vo sin
53. • At the highest point
v = vx = vy = 0
• Hence,
•
Vertically Launched Projectiles
becomes
g
• This is the time to reach the maximum height, H.
• The time taken for the projectile to return to the x-axis,
is know as the time of flight T = 2t.
vy = voy − gt
0 = vo sin− gt
t =
vo sin
54. • To find an expression for the maximum height, H
Vertically Launched Projectiles
2 2 2 2 2 2
v2
sin2
2
v sin v sin
o
− o
g 2g 2g
v sin
H = o
2g
o
y = v t −
1
gt2
g
v sin
H = vo sin o
−
g
1 v sin
2
2
g
o
H =
−
1
=
1
o
:1
=
2 2
55. • To find an expression for the maximum horizontal
distance covered, referred to as the Range, R
x = vo x t T = 2t
• Applying Trig. Double angle identity sin 2= 2sincos
•
Vertically Launched Projectiles
g g
v sin v 2
2 sin co s
R = vo cos 2 o
= o
v2
sin 2
R = o
g
56. • The trajectory; the path taken by the projectile motion
can also be derive by using the equation
• The expression shows how y and x are related
• The equation of parabola is given as
y = bx − ax2
Vertically Launched Projectiles
2
o y
y = v t −
1
gt 2
2
v cos
y = tanx −
o
y = v sin
o
x 1
v cos
−
2
g
o
x2
2v2
cos2
x
g
o
57. Example
=
A goalkeeper kicks a football with a velocity of 20.0 m/s and at
an angle of 53 degrees.
(a) How long is the ball in the air?
(b) How far away does it land?
(c) How high does it travel?
vox = vo cos
vox = 20cos53 =12.04 m/ s
voy = vo sin
voy = 20sin53 =15.97m / s
58. Example
Agoalkeeper kicks a
football with a velocity of
20.0 m/s and at an angle
of 53 degrees.
(a) How long is the ball in
the air?
What I know What I want
to know
vox=12.04 m/s t = ?
voy=15.97 m/s x = ?
y = 0 ymax=?
g = - 9.8
m/s/s
−15.97t = −4.9t2
→15.97 = 4.9t
2
oy
y = v t + 1 gt2
→ 0 = (15.97)t − 4.9t2
t = 3.26 s
59. Example
Agoalkeeper kicks a
football with a velocity of
20.0 m/s and at an angle
of 53 degrees.
(b) How far away does it
land?
What I know What I want
to know
vox=12.04 m/s t = 3.26 s
voy=15.97 m/s x = ?
y = 0 ymax=?
g = - 9.8
m/s/s
x = voxt → (12.04)(3.26) = 39.24 m
60. Example
Agoalkeeper kicks a football
with a velocity of 20.0 m/s
and at an angle of 53
degrees.
(c) How high does it travel?
CUTYOUR TIME IN HALF!
What I know What I want
to know
vox=12.04 m/s t = 3.26 s
voy=15.97 m/s x = 39.24 m
y = 0 ymax=?
g = - 9.8
m/s/s
y = (15.97)(1.63) − 4.9(1.63)2
2
oy
y = v t + 1 gt2
y = 13.01 m
63. EXERCISE 1
• Atrain changes its position (x) as a function of time (t)
as follows: x = 10 +3t2, with x m.
• (a) find the displacement of the train between t1 = 1.0 s
and t2 = 3.0 s.
• (b) find the average velocity during the same interval.
• (c) find the instantaneous velocity at the time t1 = 3.0 s
• (d) find the acceleration of the train.
64. EXERCISE 2
• The position of a projectile travelling in 2D space is
given by x(t)= (2t2 + 4t + 3)m and y(t) = (3t2 – 5t + 2)m.
calculate the magnitude and direction of the projectile’s
displacement and the average velocity between time
interval t =2 s and t = 5 s.
65. EXERCISE 3
• A stone is projected vertically upwards into the air at
40 m/s. After 3 s another stone is projected vertically
upward and it crosses the first stone at 50 m from the
ground. What is the velocity of projection of the
second stone.
66. EXERCISE 4
• An object is dropped from a height of 50m.
• (a) What is the velocity before the object hits the
ground.
• (b) what is the time required for the object to reach the
ground.