Leaning to deal with graphs

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Leaning to deal with graphs

• Time Interval, Displacement and Average Speed
• Average Velocity
• Uniform Rectilinear Motion Worldline
• Acceleration
• Velocity vs Time Graph (Constant Acceleration)
• Position vs Time for constant acceleration
• Worldline for an object moving under a constant
  acceleration
• Average velocity of a constant accelerated motion
• Displacement in a constant accelerated motion
• Gravity

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Time Interval, Displacement and Average Speed
An object is moving on a straight line for 10 s
to the right. Represent graphically this
situation.
x0 t0
x t
Time interval
Displacement
Average Speed
The time interval is a scalar. That is, it is a
quantity that can be represented by a number. A
scalar has a magnitude but not a direction.
The displacement is a vector. That is, it is a
quantity that has both size and direction. In one
dimension the direction is represented by the
sign If x gt x0, positive, and if x lt x0,
negative. Its magnitude ?x is a scalar.
The average speed is a scalar. Since, both ?t
and ?x are scalars.
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Ex. An object traveled a distance of 200 m on a
straight line to the right in 10 s and then moved
back by 5 s decreasing its average speed four
times. Find the objects displacement.
To the left!
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Average Velocity
Lets suppose that an object is moving uniformly
on a straight line. Then, it will be traveling
the same displacement in the same time intervals,
and we can define de constant velocity vector as
In one dimension the arrow can be omitted, and
the direction represented by the sign An object
moving to the right has a positive velocity,
while moving to the left will have a negative
velocity.
From the above definition, and taking into
account that ?t is a scalar we can write
Defining the initial time at t0 0
This is the equation of a straight line, with
intercept x0 and slope v. Notice that this
interpretation is valid because we are
considering a constant velocity (i.e. a constant
slope)
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Uniform Rectilinear Motion Worldline
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Ex. A person walks 90 m east in 30 s, then 10 m
west in 20 s. Find the average speed and average
velocity.
90 m
10 m
80 m
Average speed
Average velocity
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Ex. A runner runs uniformly 250 m in 25 s and
then walks back in 150 s. Find the average
velocity and the average speed for the complete
motion.
Average speed
Average velocity
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Ex. A runner runs uniformly 250 m in 25 s and
then walks back in 150 s. Represent the worldline
for the whole motion.
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Ex. Two runners run on a straight track starting
simultaneously and running in the same direction
with constant speeds of 5.50 m/s and 7.5 m/s
respectively. The slower runner starts 50 m ahead
of the faster one. When will the faster one
overtake the slower one and how far from the
starting point will they have run?
When will the fast one overtake the slower one?
How far from the starting point will they have
run?
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Acceleration
Lets suppose that an object is moving on a
straight line with a velocity which is uniformly
changing in time. Then, its velocity will be
changing by the same amount ?v in the same time
intervals ?t, and we can define de constant
acceleration vector as
In one dimension the arrow can be omitted, and
the direction represented by the sign An object
increasing its velocity has a positive
acceleration, while decreasing its velocity will
have a negative acceleration.
From the above definition, and taking into
account that ?t is a scalar we can write
Defining the initial time at t0 0
This is the equation of the instantaneous
velocity as a function of time. Its a straight
line, with intercept v0 and slope a. Notice that
this interpretation is valid because we are
considering a constant acceleration (i.e. a
constant slope)
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Velocity vs Time Graph (Constant Acceleration)
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Ex. Describe the motion represented in the graph
for the time intervals from 0 to t1 and from t1
to t2.
a gt0 and v lt 0 The acceleration and velocity are
in opposite directions ? the object will slow
down.
a gt0 and v gt 0 The acceleration and velocity
are in the same direction ? the object will
speed up.
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Ex. If the velocity of an object is zero, does
it mean that the acceleration is zero?
No, because the acceleration means change of
velocity, and the velocity can be changing from
zero to a different value, meaning that there is
an acceleration different from cero.
Ex. If the acceleration is zero, does it mean
that the velocity is zero?
No, because an object can be moving at a constant
velocity (i.e. keeping the same speed in an
straight line) and because the velocity is not
changing the acceleration is zero.
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Position vs Time for constant acceleration
The space traveled from t0 0 to t is given by
the area under the graph x-X0 A? A?
½ t (v-v0) v0 t ½ t v ½ t v0
½ t (at v0) ½ t v0 ½ at2 t
v0
X ½ at2 t v0 X0
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Worldline for an object moving under a constant
acceleration
a gt 0
a lt 0
X ½ at2 t v0 X0
The graph of this equation is a parabola, as
shown in the above figures.
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Average velocity of a constant accelerated motion
The average velocity can be considered as a
constant velocity that will make the object to
travel the same distance as the one traveled at
the corresponding variable velocity in the same
time
Note that this formula is only valid when the
acceleration is constant.
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Displacement in a constant accelerated motion
From the obtained formula
The distance traveled from t 0 to t is equal to
the area of the rectangle under
Combining this formula with v v0 at we obtain
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Summary of formulas for a const.
x ½ at2 t v0 x0
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Ex. At what time an object moving with constant
acceleration from t 0 to t tf will have a
velocity equal to the average velocity?
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Ex. 2-3 Suppose that an object at time t 0 is
at the origin, with an initial positive velocity
of v0 and is subjected to an acceleration that
varies in time as shown in the Fig. Draw the
velocity and the position-vs-time curves that
correspond to the acceleration and initial
velocity.
a
t1
t2
t
v
v0
t
t2
t1
x
Parabolic
t
t2
t1
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Gravity
The acceleration of gravity near the surface of
the Earth can be considered constant with an
approximated value of 9.8 m/s2, but its exact
value depends on location.
The acceleration of gravity is always pointing in
the downward direction. Then, if it is set up a
coordinate system with the y axis pointing up, g
will be negative in the kinematics equations.
y
v
g
x
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Ex. 2-4 A ball is thrown straight up, with an
initial velocity of 20 m/s. Find out how long it
takes the ball to reach its highest point, and
find how high the ball goes.
Data v020 m/s, a g 9.8 m/s2,
Questions t ?, y ?. when v 0
Equations
At the highest point v 0, thus
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Ex. An object accidentally falls from rest from
the ledge of a 53.0 m building. When the object
is 14.0 m above the ground, a man 2.00 m tall,
looks up and notices that the object is directly
above him. How much time, at most, does the man
have to get out of the way?
V0 0
Question t ? Time for the object to travel
the distance 12.0 m
53.0 m
V1 ?
14.0 m
2.00 m
A/
t 0.41 s since the time cannot be negative