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PHY 151: Lecture 2
 Study of Motion
 2.1 Average Velocity
 2.2 Instantaneous Velocity
 2.3 Particle under Constant Velocity
 2.4 Acceleration
 2.5 Motion Diagrams
 2.6 Particle Under Constant Acceleration
 2.7 Freely Falling Objects
 2.8 Context Connection: Acceleration
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Study of Motion - 1
• Everything in the universe is moving
• To understand the universe, we must
understand motion
• Historically, motion was the first classical
physics topic that was understood

Study of Motion - 2
• Mechanics
Branch of physics that deals with motion
• Kinematics
Description of motion
• Dynamics
What causes motion or changes motion

PHY 151: Lecture 2
Motion in One Dimension
2.1
Average Velocity

Average Velocity - 1
• Kinematics describes motion while
ignoring the agents that caused the motion
• For now, we will consider motion in one
dimension
• Along a straight line
• We will use the particle model
• A particle is a point-like object, has mass but
infinitesimal size

• Average speed is defined as
• Speed is not a vector, so has no direction
• Average velocity is a vector quantity

• The motion of a particle can be specified if its
position is known at all times
• Consider the car moving back and forth along
the x axis
• The figure is a pictorial representation of the
motion
• Imagine we take data
on the position of the
car every 10 s

Average Velocity – 4
• The graphical
representation of the
motion is a
position-time
graph
• The smooth curve is
a guess as to what
happened between
the data points

• The tabular
representation of
the motion is
shown here
• The data are
entries for
position at each
time

• Average velocity: defined as ratio of
displacement x to time interval t
• Subscript x indicates motion along x axis
• Dimensions: m/s (SI)
• x can be positive or negative

• Average velocity can be interpreted geometrically
• A straight line can be drawn between any two points on
curve
• Line forms the hypotenuse of right triangle of height x
and base t
• Slope of hypotenuse is ratio x/t
• Average velocity during time
interval t is slope of the line
joining initial and final points
on the position-time graph

• The displacement of a particle during the time
interval ti to tf is equal to the area under the
curve between the initial and final points on a
velocity-time curve

Example 2.1
Find the displacement, average velocity, and
average speed of the car between positions A
and F.
– Find the displacement of the car:
– Find the car’s average velocity:

Example 2.1
– Find the car’s average speed:
• We cannot unambiguously find
the average speed of the car
from the data in Table 2.1,
because we do not have
information about the positions
of the car between the data
points
• Assume the distance from A to B
is 22 m. The distance from B to F
is 105 m for a total o 127 m
distance.

Example 2.2
A jogger runs in a straight line, with an
average velocity of magnitude 5.00 m/s for
4.00 min and then with an average velocity
of magnitude 4.00 m/s for 3.00 min.

Example 2.2
(A) What is the magnitude of the final
displacement from her initial position?
• Find the displacement for each portion:

Example 2.2
(B) What is the magnitude of her
average velocity during this entire time
interval of 7.00 min?
• Find the average velocity for the entire
time interval:

PHY 151: Lecture 2
2.2
Instantaneous Velocity

Instantaneous Velocity - 1
• Instantaneous velocity is the limit of the
average velocity as the time interval
becomes infinitesimally short
• or as the time interval approaches zero
• The instantaneous velocity indicates what
is happening at every point of time

• The instantaneous velocity is the slope of the
line tangent to the x vs. t curve
• the green line in the figure
• The blue lines show that as t gets smaller,
they approach the green line

• The general equation for instantaneous
velocity is
• dx/dt is the derivative of x with respect to t
• The instantaneous velocity
can be positive, negative,
or zero

• Instantaneous speed: magnitude of the
instantaneous velocity vector
• Speed can never be negative

Example 2.3
The position of a particle moving along the x axis
varies in time according to the expression x = 3t2
,
where x is in meters and t is in seconds. Find the
velocity in terms of t at any time.
• The initial coordinate at time t is xi = 3t2
, the coordinate at
a later time t + t is
• Find the displacement in the time interval t:

Example 2.3
– Find the average velocity in this time interval:
– To find the instantaneous velocity,
take the limit as t approaches
zero:
– At t = 3 s, vx = 18 m/s

Example 2.4
A particle moves along the x
axis. Its position varies with
time according to the
expression x = 4t + 2t2
,
where x is in meters and t is
in seconds. Notice that the
particle moves in the
negative x direction for the
first second of motion, is
momentarily at rest at the
moment t = 1 s, and moves in
the positive x direction at
times t > 1 s.

Example 2.4
(A) Determine the displacement of the particle in
the time intervals t = 0 to t = 1 s and t = 1 s to t =
3 s.
• In the first time interval, set ti = tA = 0 and tf = tB = 1 s, find
the displacement:
• For the second time interval (t = 1 s to t = 3 s), set ti
= tB = 1 s and tf = tD = 3 s:
• These displacements can also be read directly from the
position–time graph.

PHY 151: Lecture 2
2.3
Particle Under Constant Velocity

Particle Under Constant Velocity - 1
• If the velocity of a particle is constant:
• Its instantaneous velocity at any instant is the
same as the average velocity over a given time
period:
vx = vx, avg = x/t
xf = xi + vxt
• These equations can be applied to particles
or objects that can be modeled as a particle
moving under constant velocity

Example – 2.5
• A kinesiologist is studying the biomechanics
of the human body. (Kinesiology is the study
of the movement of the human body. Notice
the connection to the word kinematics.) She
determines the velocity of an experimental
subject while he runs along a straight line at a
constant rate. The kinesiologist starts the
stopwatch at the moment the runner passes
a given point and stops it after the runner has
passed another point 20 m away. The time
interval indicated on the stopwatch is 4.0 s.

Example – 2.5
(A) What is the runner’s velocity?
• Find the constant velocity of the runner:
(B) If the runner continues his motion after the
stopwatch is stopped, what is his position after
10 s has passed?
• Find the position of the particle (runner) at time
t = 10 s:

PHY 151: Lecture 2
2.4
Acceleration

Acceleration - 1
• Acceleration is the rate of change of the
velocity
• It is a measure of how rapidly the velocity is
changing
• Dimensions are L/T2
• SI units are m/s2

Acceleration – 2
• Instantaneous acceleration is the limit of the
average acceleration as t approaches zero:
• Because vx = dx/dt, this can also be written
as

Acceleration – 3
• The slope of the
velocity vs. time graph
is the acceleration
• Positive values
correspond to where
velocity in the positive x
direction is increasing
• The acceleration is
negative when the
velocity is in the positive
x direction and is
decreasing

Acceleration – 4
• Negative acceleration does not necessarily
mean that an object is slowing down
– If the velocity is negative and the acceleration is
negative, the object is speeding up
• Deceleration is commonly used to indicate an
object is slowing down, but will not be used in
this text
– When an object’s velocity and acceleration are in
the same direction, the object is speeding up in
that direction
– When an object’s velocity and acceleration are in
the opposite direction, the object is slowing down

Example 2.6
The velocity of a particle moving along the x
axis varies according to the expression
vx = 40  5t2
where vx is in meters per second and t is in
seconds.

Example 2.6
(A) Find the average acceleration in the time
interval t = 0 to t = 2.0 s.
• Find the velocities at ti = tA = 0 and tf = tB = 2.0 s:
• Find the average acceleration in the specified
time interval:

Example 2.6
(B) Determine the acceleration at t = 2.0 s.
• Knowing that the initial velocity at any time t is vxi
= 40  5t2
, find the velocity at any later time t +
t:
• Find the change in velocity over the time interval
t:

Example 2.6
– Divide by t and take the limit:
– Substitute t = 2.0 s:

PHY 151: Lecture 2
2.5
Motion Diagrams
Skipped

PHY 151: Lecture 2
2.6
Particle Under Constant Acceleration
Skipped Some

Particle Under Constant Acceleration - 1
• If a particle has constant acceleration, we can
use the following equations:

Example 2.7
A jet lands on an aircraft carrier at a speed of
140 mi/h ( 63 m/s).
(A) What is its acceleration (assumed constant)
if it stops in 2.0 s due to an arresting cable that
snags the jet and brings it to a stop?
• Find the acceleration of the jet, modeled as a
particle:

Example 2.7
• (B) If the jet touches down at position xi = 0,
what is its final position?

Example 2.8
A car traveling at a constant speed of 45.0 m/s
passes a trooper on a motorcycle hidden
behind a billboard. One second after the
speeding car passes the billboard, the trooper
sets out from the billboard to catch the car,
accelerating at a constant rate of 3.00 m/s2
.
How long does it take her to overtake the car?

Example 2.8
• Find the position at any time t:
• The trooper starts from rest at tB = 0 and
accelerates at ax = 3.00 m/s2
away from the origin.
Find her position at any time t:

Example 2.8
• Set the positions of the car and trooper equal to
represent the trooper overtaking the car at
position C:
• Rearrange the quadratic equation:

Example 2.8
• Solve the equation for time:
• Evaluate the solution (choose the positive root):

1-Dimension Horizontal Motion
Example 1
• Basketball player starts from rest and accelerates uniformly to speed 6.0 m/s in 1.5 s
• What distance does the player run?
 vf = vi + at
 xf = xi + vit + ½at2
 vf
2
= vi
2
+ 2a(xf – xi)
 xi = 0 xf = ?
 vi = 0 vf = 6.0
 a = ? t = 1.5
 6=0+a(1.5) Equation has 1 unknown, a (1st
)
 xf=0+0(1.5)+½a(1.5)2
Equation has 2 unknowns, xf, a (2nd
)
 62
=02
+2a(xf–0) Equation has two unknowns, xf, a
 6=0+a(1.5); a = 6/1.5 = 4 m/s2
 xf=0+0(1.5)+½(4)(1.5)2
= 4.5 m/s

Example 2a
• Car accelerates from rest at rate of 2.0 m/s2
for 5.0 s
• (a) What is the speed of the car at the end of that time?
 vf = vi + at
 vf
2
= vi
2
+ 2a(xf – xi)
 xi = 0 xf = ?
 vi = 0 vf = ?
 a = 2.0 t = 5.0
 vf=0+2(5) Equation has 1 unknown, vf
 xf=0+0(5)+½(2)(5)2
Equation has 1 unknown, xf
 vf
2
=02
+2(2)(xf–0) Equation has two unknowns, xf, vf
 vf=0+2(5)=10 m/s

Example 2b
• Car accelerates from rest at rate of 2.0 m/s2
for 5.0 s
• (b) How far does the car travel in this time?
 vf = vi + at
 vf
2
= vi
2
+ 2a(xf – xi)
 xi = 0 xf = ?
 vi = 0 vf = 10
 a = 2.0 t = 5.0
 10=0+2(5) Can’t use Equation
 xf=0+0(5)+½(2)(5)2
Equation has 1 unknown, xf
 102
=02
+2(2)(xf–0) Equation has 1 unknown, xf
 xf=0+0(5)+½(2)(5)2
= 25 m

Example 3a
• A car traveling at 15 m/s stops in 35 m
• (a) What is the acceleration?
 vf = vi + at
 vf
2
= vi
2
+ 2a(xf – xi)
 xi = 0 xf = 35
 vi = 15 vf = 0
 a = ? t = ?
 0=15+at Equation has 2 unknowns, a, t
 35=0+15t+½(a)(t)2
Equation has 2 unknowns, a, t
 02
=152
+2a(35–0) Equation has 1 unknown, a
 02
=152
+2a(35–0)
 a = -225/70 = -3.2 m/s2

Example 3b
• A car traveling at 15 m/s stops in 35 m
• (b) What is time during this deceleration until car stops?
 vf = vi + at
 vf
2
= vi
2
+ 2a(xf – xi)
 xi = 0 xf = 35
 vi = 15 vf = 0
 a = -3.2 t = ?
 0=15-3.2t Equation has 1 unknown, t
 35=0+15t+½(-3.2)(t)2
Equation has 1 unknown, t
 02
=152
+2(-3.2)(35–0) Can’t use equation
 0=15-3.2t
 t = -15/-3.2 = 4.67 s

Example 4
• A plane accelerates at 8 m/s2
on a runway that is 500 m long
• The take off speed of the plane is 80 m/sec
• Can the plane takeoff?
 vf = vi + at
 vf
2
= vi
2
+ 2a(xf – xi)
 xi = 0 xf = 500
 vi = 0 vf = ?
 a = 8 t = ?
 vf=0+8t Equation has 2 unknowns, vf, t
 500=0+0t+½(8)(t)2
 vf
2
=02
+2(8)(500–0) Equation has 1 unknown, vf
 vf
2
=02
+2(8)(500–0) = 500(16)
 vf=sqrt(500(16) = 89.4 m/s
• The plane can takeoff

Example 5
• A plane accelerates at 8 m/s2
• The take off speed of the plane is 80 m/sec
• What is minimum length of run way for plane to reach take
off speed?
 vf = vi + at
 vf
2
= vi
2
+ 2a(xf – xi)
 xi = 0 xf = ?
 vi = 0 vf = 80
 a = 8 t = ?
 80=0+8t Equation has 1 unknown, t
 xf=0+0t+½(8)(t)2
Equation has 2 unknowns, xf, t
 802
=02
+2(8)(xf–0) Equation has 1 unknown, xf
 802
=02
+2(8)(xf–0)
 xf=6400/16 = 400 m

Example 6 Bull
• Bull runs 8 m/sec
• A boy at rest has a head start of 12 m
• When he sees the bull he accelerates at 2 m/s2
• Does the bull catch the boy?
vf = vi + at
xf = xi + vit + ½at2
vf
2
= vi
2
+ 2a(xf – xi)
xi = 0 xf = ?
vi = 8 vf = 8
a = 0 t = ?
8=8+0t Can’t use this equation
xf=0+8t+½(0)(t)2
Equation has 2 unknowns xf, t
82
=82
+2(0)(xf–0) Can’t use this equation

Example 6 Boy
• A boy at rest has a head start of 12 m
 vf = vi + at
 vf
2
= vi
2
+ 2a(xf – xi)
 xi = 12 xf = ?
 vi = 0 vf = ?
 a = 2 t = ?
 vf=0+2t Equation has 2 unknowns, vf, t
 xf=12+0t+½(2)(t)2
 vf
2
=02
+2(2)(xf–12) Equation has 2 unknowns, xf, vf

Example 6 Bull / Boy
• A boy at rest has a head start of 12m
 xf=0+8t+½(0)(t)2
(from Bull equations)
 xf=12+0t+½(2)(t)2
(from Boy equations)
 Bull catches boy when their xf are equal
 8t=12+t2
 t2
-8t+12=0
 Factor
 (t-2)(t-6) = 0
• Bull catches boy at 2 s and 6 s
• 2 s is bull catching up to the boy
• 6 s is the accelerating boy catching up to the bull

Example 7 Car
• Car travels at 45.0 m/s passes a trooper
• One second later the trooper sets out to catch the car accelerating at
3.00 m/s2
• How long does it take the trooper to overtake the car?
• Let t = 0 be 1 second after the car passes the trooper
• The cars position is then at xi = 45 m
 vf = vi + at
 vf
2
= vi
2
+ 2a(xf – xi)
 xi = 45 xf = ?
 vi = 45 vf = 45
 a = 0 t = ?
 45=45+0t Can’t use this equation
 xf=45+45t+½(0)(t)2
 452
=452
+2(0)(xf–0) Can’t use this equation

Example 7 Trooper
• One second later the trooper sets out to catch the car
accelerating at 3.00 m/s2
vf = vi + at
xf = xi + vit + ½at2
vf
2
= vi
2
+ 2a(xf – xi)
xi = 0 xf = ?
vi = 0 vf = ?
a = 3 t = ?
vf=0+3t Equation has 2 unknowns, vf, t xf=0+0t+½(3)(t)2
vf
2
=02
+2(3)(xf–0) Equation has 2 unknowns, xf, vf

Example 7 Car / Trooper
• One second later the trooper sets out to catch the car
accelerating at 3.00 m/s2
xf=45+45t+½(0)(t)2
(from Car equations)
xf=0+0t+½(3)(t)2 (
from Boy equations)
Trooper catches car when their xf are equal
45+45t = 1.5t2
t2
-30t-30=0
Solve using quadratic equation
t = 30.969 s rounded to 31.0 s

PHY 151: Lecture 2
2.7
Freely Falling Objects

Freely Falling Objects - 1
• Galileo performed many systematic
experiments on objects moving on inclined
planes
• With careful measurements, he showed
displacement of object starting from rest is
proportional to the square of the time interval

Freely Falling Objects - 2
• A freely falling object is any object moving
freely under the influence of gravity alone
(neglecting air resistance)
• All objects fall with the same acceleration near
Earth’s surface, 9.80 m/s2
downward
• A freely falling object can be:
– Dropped (released from rest)
– Thrown downward
– Thrown upward
• An object in free fall can be
modeled as a particle under
constant acceleration

Example 2.9
• A stone thrown from the top of
a building is given an initial
velocity of 20.0 m/s straight
upward. The stone is launched
50.0 m above the ground, and
the stone just misses the edge
of the roof on its way down.

Example 2.9
(A) Using t𝖠 = 0 as the time the stone leaves the
thrower’s hand at position 𝖠, determine the time
at which the stone reaches its maximum height.
– Calculate the time at which the stone reaches its
maximum height:
– Substitute numerical values:

Example 2.9
(B) Find the maximum height of the stone.
• Set yA = 0 and substitute the time from part (A) to
find the maximum height:

Example 2.9
(C) Determine the velocity of the stone when it
returns to the height from which it was thrown.
• Choose the initial point where the stone is
launched and the final point when it passes this
position coming down:

Example 2.9
(D) Find the velocity and position of the stone
at t = 5.00 s.
• Calculate the velocity:
• Find the position at t = 5.00 s:

1-Dimension Free Fall - Example 1
• A bomb is dropped from 6000m
• With what speed does it hit the ground?
 vf = vi + at
 yf = yi + vit + ½at2
 vf
2
= vi
2
+ 2a(yf – yi)
 yi = 6000 yf = 0
 vi = 0 vf = ?
 a = -9.8 t = ?
 vf = 0 – 9.8t Equation has 2 unknowns vf, t
 0 = 6000 + 0t + ½ (-9.8)t2
 vf
2
= 02
+ 2(-9.8) (0 – 6000) Equation has 1 unknown, vf
 vf
2
= 02
+ 2(-9.8)(-6000)
 vf = sqrt(117600) = -343 m/s

1-Dimension Free Fall – Example 2a
• A student drops a ball from the top of a tall building
• It takes 2.8 s for the ball to reach the ground
• (a) What is the height of the building?
 vf = vi + at
 vf
2
= vi
2
+ 2a(yf – yi)
 yi = ? yf = 0
 vi = 0 vf = ?
 a = -9.8 t = 2.8
 vf = 0 – 9.8(2.8) Equation has 1 unknown, vf
 0 = yi + 0(2.8) + ½(-9.8)(2.8)2
Equation has 1 unknown, yi
 vf
2
= 02
+ 2(-9.8) (0 – yi) Equation has 2 unknowns, yi, vf
 0 = yi + (1/2)(-9.8)(2.8)2
 yi = 38.4 m

1-Dimension Free Fall – Example 2b
• A student drops a ball from the top of a tall building
• It takes 2.8 s for the ball to reach the ground
• (b) What was the ball’s velocity just before hitting the ground?
 vf = vi + at
 vf
2
= vi
2
+ 2a(yf – yi)
 yi = 38.4 yf = 0
 vi = 0 vf = ?
 a = -9.8 t = 2.8
 0 = 38.4 + 0(2.8) + ½(-9.8)(2.8)2
Can’t use equation
 vf
2
= 02
+ 2(-9.8) (0 – 38.4) Equation has 1 unknown, vf
 vf = 0 – 9.8(2.8) = -27.4 m/s (- means downward)

1-Dimension Free Fall – Example 3
• Boy throws stone straight upward with an initial velocity of 15 m/s
• What maximum height will stone reach before falling back down?
• At the maximum height vy = 0
 vf = vi + at
 vf
2
= vi
2
+ 2a(yf – yi)
 yi = 0 yf = ?
 vi = 15 vf = 0
 a = -9.8 t = ?
 0 = 15 – 9.8t Equation has 1 unknown, t
 yf = 0 + 15t + ½(-9.8)t2
Equation has 2 unknown, yf, t
 02
= 152
+ 2(-9.8) (yf – 0) Equation has 1 unknown, yf
 02
= 152
+ 2(-9.8) (yf – 0)
 yf = -152
/2/-9.8 = 11.48 m

1-Dimension Free Fall – Example 4
• Ball thrown upwards at 40m/s. Calculate time to reach 35m
 vf = vi + at
 vf
2
= vi
2
+ 2a(yf – yi)
 yi = 0 yf = 35
 vi = 40 vf = ?
 a = -9.8 t = ?
 vf = 40 – 9.8t Equation 2 unknown vf,t (2nd
)
 35 = 0 + 40t + ½(-9.8)t2
 vf
2
= 402
+ 2(-9.8) (35 – 0) Equation 1 unknown, vf (1st)
 vf
2
= 402
+ 2(-9.8) (35 – 0) = 914
 vf = +30.2 or -30.2
 30.2 = 40 – 9.8t
 t = (30.2 – 40) / (-9.8) = 1.0 s (time on the way up)

1-Dimension Free Fall
Example 5 Ball 1
• A ball is dropped from rest
• Four seconds later, a second ball is thrown down at 50m/s
Calculate when and where the two balls meet
• Let t = 0 at 4 seconds. Ball height y=½(-g)t2
=½(-9.8)(16) =
-78.4 m. Speed v = (-g)t = -9.8(4) = -39.2 m/s
 vf = vi + at
 vf
2
= vi
2
+ 2a(yf – yi)
 yi = -78.4 yf = ?
 vi = -39.2 vf = ?
 a = -9.8 t = ?
 vf = -39.2 – 9.8t Equation has 2 unknowns, vf, t
 yf = -78.4 – 39.2t + ½(-9.8)t2
Equation 2 unknowns yf, t
 vf
2
= (-29.2)2
+2(-9.8) (yf – (-78.4)) Equation 2 unknowns, yf,
vf

Example 5 Ball 2
• Four seconds later, a second ball is thrown down at 50m/s
Calculate when and where the two balls meet
 vf = vi + at
 vf
2
= vi
2
+ 2a(yf – yi)
 yi = 0 yf = ?
 vi = -50 vf = ?
 a = -9.8 t = ?
 vf = -50–9.8t Equation has 2 unknown, vf, t
 yf = 0-50t+½(-9.8)t2
Equation has 2 unknowns, yf, t
 vf
2
= 502
+2(-9.8) (yf – 0) Equation has 2 unknowns, yf, vf

Example 5 Balls 1 and 2
• Four seconds later, a second ball is thrown
down at 50m/s
• Calculate when and where two balls meet
yf = -78.4 – 39.2t + ½(-9.8)t2
(from ball 1)
yf = 0-50t+½(-9.8)t2
(from ball 2)
Balls meet when positions are the same 2
-78.4-39.2t+ ½ (-9.8)t2
= -50t + ½(-9.8)t2
0 = 10.8t -78.4
t = -78.4 / -10.8 = 7.3 s

1-Dimension Free Fall – Example 6a
• From tower 100 m high, ball is thrown up with speed of 40 m/s
• (a) How high does it rise?
 vf = vi + at
 vf
2
= vi
2
+ 2a(yf – yi)
 yi = 100 yf = ?
 vi = 40 vf = 0
 a = -9.8t = ?
 0 = 40 – 9.8t Equation has 1 unknown, t
 yf = 100 + 40t + ½(-9.8)t2
Equation has 2 unknowns, yf, t
 02
= 402
+2(-9.8)(yf–100) Equation has 1 unknown, yf
 02
= 402
+2(-9.8)(yf–100)
 yf = (-1600/2/-9.8) + 100 = 181.6 m

1-Dimension Free Fall – Example 6b
• (b) When does it hit ground?
 vf = vi + at
 vf
2
= vi
2
+ 2a(yf – yi)
 yi = 100 yf = 0
 vi = 40 vf = ?
 a = -9.8 t = ?
 vf = 40 – 9.8t Equation has 2 unknowns, vf, t
 0 = 100 + 40t + ½(-9.8)t2
 vf
2
= 402
+2(-9.8)(0–100) Equation has 1 unknown, vf
 0 = 100 + 40t + ½(-9.8)t2
 -4.9t2
+ 40t + 100 = 0 (Use quadratic equation)
 t = 10.2 s

1-Dimension Free Fall – Example 6c
• (c) How fast is it moving at the ground ?
 vf = vi + at
 vf
2
= vi
2
+ 2a(yf – yi)
 yi = 100 yf = 0
 vi = 40 vf = ?
 a = -9.8 t = 10.2
 0 = 100+40(10.2)+½(-9.8)(10.2)2
Can’t use equation
 vf
2
= 402
+2(-9.8)(0–100) Equation has 1 unknown, vf
 vf
2
= 402
+2(-9.8)(0–100) = 1600 + 1960 = 3560
 vf = -59.7 m/s

PHY 151: Lecture 2
2.8
Context Connection: Acceleration Required
by Consumers
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