### 5-11-10 1-D Kinematics

```1-D Kinematics
• Kinematics
– Kinematics is the branch of mechanics that describes
the motion of objects without necessarily discussing
what causes the motion.
– 1-Dimensional Kinematics (or 1 - Dimensional motion)
refers to motion in a straight line.
• Distance
– The total length of the path traveled by an object is
called distance.
– “How far have you walked?” is a typical distance
question.
– The SI unit of distance is the meter (m).
• Displacement (Dx)
– The change in the position of a particle is called
displacement.
– ∆ is a Greek letter used to represent the words “change
in”. ∆x therefore means “change in x”. It is always
calculated by final value minus initial value.
– “How far are you from home?” is a typical displacement
question.
– The SI unit for displacement is the meter. Calculation of
displacement: ∆x = x - x
f
i
• Distance vs Displacement
distance
displacement
50m
distance and displacement.
100m
• Another example distance displacement
• Walk east 100m
Distance = 100m
• Turn and walk west 100m
Distance = 200m
Displacement = 0m
Displacement = 100m
• Questions
• Does the odometer in your car measure distance or
displacement?
– Distance
• Can you think of a circumstance in which it
measures both distance and displacement?
– If you go one direction and stop it will give both
distance and displacement
• If Dx is the displacement of a particle, and d is the
distance the particle traveled during that
displacement, which of the following is always a
true statement?
a) d = |∆x|
b) d < |∆x|
c) d > |∆x|
d) d > |∆x|
e) d < |∆x|
• A particle moves from x = 1.0 meter to x = -1.0
meter.
• What is the distance d traveled by the particle?
2m
-3 m -2 m
-1 m
0m
1m
2m
3m
• What is the displacement of the particle?
-2m, must show direction
• You are driving a car on a circular track of diameter
40 meters. After you have driven around 2 ½ times,
how far have you driven, and what is your
displacement?
C = 2πr
C = 2π∙20m
40m
C = 125.66m
Distance = 2.5∙125.66m = 314.16m
Displacement = 40m
• Average Speed
– Average speed describes how fast a particle is moving.
The equation is:
save = d/t
– where:
save = average speed
d = distance
Average speed is
always a positive
number.
∆t = elapsed time
The SI unit of speed is the m/s
• Average Speed
• Average velocity describes how fast the
displacement is changing. The equation is:
vave = ∆x/ ∆t
– where:
vave = average velocity
∆x = displacement
Average velocity is
+ or – depending on
direction.
∆t = elapsed time
The SI unit of speed is the m/s
• How long will it take the sound of the starting gun
to reach the ears of the sprinters if the starter is
stationed at the finish line for a 100 m race?
Assume that sound has a speed of about 340 m/s.
∆t = ?
∆x = 100m
vavg = 340 m/s
vave = ∆x/ ∆t ∆t= ∆x/ vave
∆t= 100m / 340 m/s = 0.294 s
• You drive in a straight line at 10 m/s for 1.0 km, and
then you drive in a straight line at 20 m/s for
another 1.0 km. What is your average velocity?
1. ∆t = ?
∆x = 1km = 1000m
vavg = 10 m/s
vave = ∆x/ ∆t ∆t= ∆x/ vave
∆t= 1000m / 10 m/s = 100 s
2.
∆t = ? ∆x = 1km = 1000m
vavg = 20 m/s
vave = ∆x/ ∆t ∆t= ∆x/ vave
∆t= 1000m / 20 m/s = 50 s
3. ∆x = 2000m
∆t total= 150s
vavg = ? m/s
vave = ∆x/ ∆t = 2000 m /150s = 13.3 m / s
• Qualitative Demonstrations
1) Demonstrate the motion of a particle that has an
average speed and an average velocity that are both
zero.
2) Demonstrate the motion of a particle that has an
average speed and an average velocity that are both
nonzero.
3) Demonstrate the motion of a particle that has an
average speed that is nonzero and an average
velocity that is zero.
4) Demonstrate the motion of a particle that has an
average velocity that is nonzero and an average
speed that is zero.
• You are a particle located at the origin. Demonstrate how
you can move from x = 0 to x = 5.0 and back with an
average speed of 0.5 m/s.
d = 10m t = ? s s= 0.5 m / s
t= d/save
save = d/t
t= 10 m / 0.5 m/s = 20 s
Cover the total
distance in 20s
• What the particle’s average velocity for the above
demonstration?
– 0 m/s because velocity takes into account the is a sign
change average velocity
– in the positive x is +0.5 m/s
– in the negative x is – 0.5 m/s
• Graphical Problem
x (m)
0m
t (s)
Demonstrate the motion of this particle
• Graphical Problem
x(m)
0m
t (s)
Demonstrate the motion of this particle
• Graphical Problem
x (m)
A
B
∆x
∆t
0m
t (s)
vave = ∆x/ ∆t
What physical feature of the graph gives
the constant velocity from A to B?
• Graphical Problem: Determine the average velocity
from the graph.
x (m)
B
2.0
A
∆x
1.0
∆t
0
-1.0
t (s)
2.0
4.0
6.0
-2.0
vave = 1 m/ 3 s = 1 / 3 m/ s
• Graphical Problem
x (m)
0m
t (s)
Demonstrate the motion of these two particle
• Graphical Problem
X (m)
0m
t (s)
Demonstrate the motion of these two particle
• Graphical Problem
X (m)
0m
t (s)
What kind of motion does this graph represent?
• Graphical Problem
x (m)
A
B
∆x
0m
∆t
T (s)
vave = ∆x/ ∆t
Can you determine average velocity from the
time at point A to the time at point B from the
raph?
• Graphical Problem: Determine the average velocity
between 1 and 4 seconds.
x (m)
∆x
2.0
∆t
1.0
0
-1.0
t (s)
2.0
4.0
6.0
-2.0
vave = ∆x/ ∆t
vave = 1 m / 4 s = 0.25 m/s
• Instantaneous Velocity
– The velocity at a single instant in time.
– If the velocity is uniform, or constant, the instantaneous
velocity is the same as the average velocity.
– If the velocity is not constant, than the instantaneous
velocity is not the same as the average velocity, and we
must carefully distinguish between the two.
• Instantaneous Velocity
X
vins = ∆x/ ∆t
∆t
0m
∆x
t
Draw a tangent line to the curve at B. The slope
of this line gives the instantaneous velocity at
that specific time.
• Determine the instantaneous velocity at 1 second.
X
∆x
2.0
∆t
1.0
0
t
-1.0
2.0
4.0
6.0
-2.0
vave = ∆x/ ∆t vave = 1 m/ 1 s = 1 m/s
• Acceleration (a)
– Any change in velocity over a period of time is called
acceleration.
– The sign (+ or -) of acceleration indicates its direction.
– Acceleration can be…
• speeding up
• slowing down
• turning
• Questions
• If acceleration is zero, what does this mean about
the motion of an object?
it means the velocity is constant at 0 m/s or
some velocity greater than 0 m/s
The object has a constant speed, and it is
moving in a straight line
• Is it possible for a racecar circling a track to have
zero acceleration?
No because the race car is changing direction. When an object
changes direction it changes velocity, so it has an acceleration
present.
• Uniform (Constant) Acceleration
– In Physics B, we will generally assume that acceleration
is constant.
– With this assumption we are free to use this equation:
a = ∆v / ∆t
– The SI unit of acceleration is the m/s2.
• Acceleration in 1-D Motion has a sign!
– If the sign of the velocity and the sign of the
acceleration is the same, the object speeds up.
– If the sign of the velocity and the sign of the
acceleration are different, the object slows down.
Qualitative Demonstrations
1) Demonstrate the motion of a particle that has
zero initial velocity and positive acceleration.
2) Demonstrate the motion of a particle that has
zero initial velocity and negative acceleration.
3) Demonstrate the motion of a particle that has
positive initial velocity and negative acceleration.
4) Demonstrate the motion of a particle that has
negative initial velocity and positive acceleration.
• Practice Problem: A 747 airliner reaches its takeoff
speed of 180 mph (80.5 m/s) in 30 seconds. What is
its average acceleration?
a = ? mph/s ∆v = vf – vi = 180 mph – 0 mph = 180 mph
∆t = tf – ti = 30 s – 0 s = 30 s
a = ∆v / ∆t
a = 180mph / 30 s = 6mph / s
a = ? m/s2 ∆v = vf – vi = 80.5 m / s– 0 m / s = 80.5 m / s
∆t = tf – ti = 30 s – 0 s = 30 s
a = ∆v / ∆t
a = 80.5 m / s / 30 s= 2.683 m / s2
• A horse is running with an initial velocity of 11 m/s,
and begins to accelerate at –1.81 m/s2. How long
does it take the horse to stop?
a = -1.81 m/s2
∆v = vf – vi = 0 m /s – 11 m /s = -11 m /s
∆t = ? s
∆t = ∆v /a
∆t = -11 m /s / -1.81 m / s2 = 6.1 s
• Graphical Problem
v (m/s)
0m
T (s)
Demonstrate the motion of this particle. Is it accelerating?
• Graphical Problem
v (m/s)
0m
T (s)
Demonstrate the motion of this particle. Is it accelerating?
• Graphical Problem
v (m /s)
A
B
∆v
∆t
0m
t (s)
a = ∆v/ ∆t
What physical feature of the graph gives
the acceleration from A to B?
• Graphical Problem: Determine the acceleration
from the graph.
v (m/s)
2.0
∆v
1.0
∆t
0
-1.0
t (s)
2.0
4.0
6.0
-2.0
a= ∆v/ ∆t
a = 1 m /s / 1 s = 1 m/s2
• Graphical Problem: Determine the displacement of
the object from 0 to 4 seconds..
Area of a triangle = ½ b ∙h
v (m/s)
2.0
h
1.0
0
-1.0
2.0
b
t (s)
4.0
6.0
h
-2.0
b Atriangle = ½ b ∙h = ½ ∙ 2 ∙2 = 2 m
Atriangle = ½ b ∙h = ½ ∙ 2 ∙2 = - 2 m
disp = 2 m – 2m = 0 m
• Need to do this example below
• 5-14-10 1-D kinematics slide 41.xlsx
x (m) 0
t (s) 0
Pinewood Derby
2.3 9.2 20.7 36.8 57.5
1.0 2.0 3.0 4.0 5.0
• On your graph paper, do the following.
a) Draw a position vs time graph for the car.
b) Draw tangent lines at three different points on the
curve to determine the instantaneous velocity at all
three points.
c) On a separate graph, draw a velocity vs time graph using
the instantaneous velocities you obtained in the step
above.
d)From your velocity vs time graph, determine the
acceleration of the car.
distance - time graph
70
60
50
40
x (m)
30
20
10
0
0
1
2
3
t (s)
4
5
6
• Uniformly Accelerating Objects
•You see the
car move faster
and faster as
time elapses.
• For each
successive ∆t,
∆x is bigger.
• The velocity
vs time graph
reflects the
increasing
velocity
What are the signs of velocity and acceleration? How is speed
changing?
This object is moving with positive
velocity and acceleration. Speed is
increasing.
This object is moving in the negative
velocity and acceleration. Speed is
increasing.
This object is moving with negative
velocity and positive acceleration.
Speed is decreasing.
• Draw Graphs for Stationary Particles
x
v
a
0m
position
vs time
t
0m/s
position
vs time
t
0 m / s2
position
vs time
t
• Draw Graphs for Constant Non-zero Velocity
x
v
a
0m
position
vs time
t
0m/s
position
vs time
t
0 m / s2
position
vs time
t
• Draw Graphs for Constant Non-zero Acceleration
increasing and positive velocity
v
x
a
0m
position
vs time
t
0m/s
position
vs time
t
0 m / s2
position
vs time
t
• Draw Graphs for Constant Non-zero Acceleration
decreasing and negative velocity
v
x
a
0m
position
vs time
t
0m/s
position
vs time
t
0 m / s2
position
vs time
t
• Draw Graphs for Constant Non-zero Acceleration
increasing and negative velocity
v
x
a
0m
position
vs time
t
0m/s
position
vs time
t
0 m / s2
position
vs time
t
• Draw Graphs for Constant Non-zero Acceleration
decreasing and negative
velocity
v
x
a
0m
t
position
vs time
0 m / s2
0m/s
t
position
vs time
position
vs time
t
• Kinematic Equations for uniformly accelerating
objects
v = v0 + at
On the AP equation sheet
x = x0 + 1/2at2
x = v0t + 1/2at2
x = x0 + v0t + 1/2at2 On the AP equation sheet
v2 = v02 + 2a(∆x)
v2 = v02 + 2a(x – x0) On the AP equation sheet
• What must a particular Olympic sprinter’s
acceleration be if he is able to attain his maximum
speed in ½ of a second?
v0 = 0 m/ s
t=½s
v = v0 + at v = at
a=v/t
a = v / (1/2)
a = 2v
• A plane is flying in a northwest direction when it lands,
touching the end of the runway with a speed of 130 m/s. If
the runway is 1.0 km long, what must the acceleration of
the plane be if it is to stop while leaving ¼ of the runway
remaining as a safety margin?
t
unk
∆x
v
V0
0 m/ s 130 m/ s 750 m
v2 = v02 + 2a(∆x)
0 = v02 + 2a(∆x)
a = v02 / (2(∆x))
a = (130 m/s)2 / (2(750m)) = - 11.23 m /s2
• On a ride called the Detonator at Worlds of Fun in Kansas
City, passengers accelerate straight downward from 0 to 20
m/s in 1.0 second.
a) What is the average acceleration of the passengers on
this ride?
t
v
v
∆x
a
0
1.0s 20 m/ s
0 m/ s
v = v0 + at v = 0 + at
v = at
a = 20 m/s / 1s = 20 m/s2
?
? m/s2
a=v/t
b) How fast would they be going if they accelerated for an
additional second at this rate? t = 2 s a = 20 m/s2
v = 0 + at
v = at
v = 20 m/s2 ∙ 2 s = 40 m /s
c) Sketch approximate x-t, v-t and a-t graphs for this
ride.
v
x
a
0m
position
vs time
t
0 m / s2
0m/s
t
position
vs time
t
position
vs time
• Air bags are designed to deploy in 10 ms. Estimate
the acceleration of the front surface of the bag as it
t = 2 ms = 0.002 s
expands.
a=?
v = v0 + at
v0 = 0 m / s
v = at
a = v / 0.002s = 500 v
• You are driving through town at 12.0 m/s when suddenly a
ball rolls out in front of you. You apply the brakes and
decelerate at 3.5 m/s2.
a) How far do you travel before stopping?
t
v
v0
∆x
a
unk 0 m/ s
12 m/ s unk
-3.5 m/s2
v2 = v02 + 2a(∆x) 0 = v02 + 2a(∆x)
(∆x) = - v02 / (2a)= - (12 m / s)2 / (2∙ - 3.5 m/s2) = 20.57
m
b) When you have traveled only half the stopping distance,
∆x = 20.57 / 2 = 10.29 m
v2 = v02 + 2a(∆x)
v = √((12 m/s)2 + 2 ∙ -3.5 m/s2 ∙ (10.29 m)) = 8.49 m / s
• Practice Problem -- continued
c) How long does it take you to stop?
t
?s
v
0 m/ s
v0
12 m/ s
v = v0 + at 0 = v0 + at
∆x
a
20.57 m -3.5 m/s2
at = v0
t = -v0 / a = -(12 m/s) /- 3.5 m/s2 = 3.43 s
• Practice Problem -- continued
d) Draw x vs t, v vs t, and a vs t graphs for this.
v
x
a
0m
position
vs time
t
0m/s
position
vs time
t
0 m / s2
position
vs time
t
• Free Fall
• Free fall is a term we use to indicate that an object
is falling under the influence of gravity, with gravity
being the only force on the object. Air resistance is
considered negligible.
• Gravity accelerates the object toward the earth
the entire time it rises, and the entire time it falls.
• Freefall is like any other type of accelerated
motion; it’s just that the acceleration is pretty well
defined. The acceleration due to gravity near the
surface of the earth has a magnitude of 9.8 m/s2.
• The direction of this acceleration is DOWN.
• You drop a ball from rest off a 120 m high cliff. Assuming air
resistance is negligible,
a) how long is the ball in the air?
t
v
v0
∆y
a
? s ? m/ s
0 m/ s
-120 m -9.8 m/s2
y = y0 + 1/2at2 y0 = 0m y = 0 + 1/2at2
t = √ (2y / a)
t = √ (2 ∙ -120m/ -9.8 m/s2) = 4.94 s
Exact same
Or
equation
y = v0t + 1/2at2
t = √ (2y / a)
y = 0 t + 1/2at2
• You drop a ball from rest off a 120 m high cliff. Assuming air
resistance is negligible,
b) what is the ball’s speed and velocity when it strikes the
ground at the base of the cliff?
t
v
v0
∆y
a
4.94s ? m/ s
0 m/ s
120 m
-9.8 m/s2
v = 0 + at
v = v0 + at
v = at = - 9.8 m/s2 ∙ 4.94s = - 48.12 m / s
Velocity = - 48.12 m / s
speed = 48.12 m / s
• You drop a ball from rest off a 120 m high cliff. Assuming air
resistance is negligible,
c) sketch approximate x-t, v-t, a-t graphs for this
situation.
v
x
0m
position
vs time
t
0m/s
position
vs time
a
t
0 m / s2
position
vs time
t
• Symmetry in Free Fall
– When something is thrown straight upward under the
influence of gravity, and then returns to the thrower, this
is very symmetric.
– The object spends half its time traveling up; half
traveling down.
– Velocity when it returns to the ground is the opposite of
the velocity it was thrown upward with.
– Acceleration is 9.8 m/s2 and directed DOWN the entire
time the object is in the air!
Objects Launched from the Ground
Ex: water rocket
**Force of water accelerates rocket off the ground. Once the water is gone, gravity takes over:
3s;
v=0m/s, a=9.8m/s2
2s;
4s;
v=9.8m/s
v=-9.8m/s
a=-9.8m/s2
a=-9.8m/s2
1s;
5s;
v=19.6m/s
v=-19.6m/s
a=- a=-9.8m/s2
a=-9.8m/s2
0s;
vi=29.4m/s
a=-9.8m/s2
6s;
vf=-29.4m/s
a=-9.8m/s2
• If object is thrown up, the upward downward
velocities and displacements are equal and
opposite.
vtop = 0 m/s, atop = -9.8 m/s2
v1 > vtop, a = -9.8 m/s2
v2 > v1, a = -9.8 m/s2
v3 > v2, a = -9.8 m/s2
• You throw a ball straight upward into the air with a velocity
of 20.0 m/s, and you catch the ball some time later.
a) How long is the ball in the air?
t
v
v0
∆y
a
?s
0 m/ s
20 m/ s ? m
-9.8 m/s2
v = v0 + at 0 = v0 + at
at = -v0
t = -v0 / a = - (20 m/s) /- 9.8 m/s2 = 2.04 s
b) How high does the ball go?
v2 = v02 + 2a(∆y) 0 = v02 + 2a(∆y)
(∆y) = -v02 / (2a) = - (20 m / s)2 / (2∙ - 9.8 m/s2) = 7.35 m
• Practice Problem -- continued
c) What is the ball’s velocity when you catch it?
• - 20 m/s, due to the symmetry of the ball motion.
d) Sketch approximate x-t, v-t, a-t graphs for this situation.
v
x
0m
position
vs time
t
0m/s
position
vs time
a
t
0 m / s2
position
vs time
t
• Reflex Testing
– Who’s has the quickest reflexes in Physics B? We’re
dying to know.
– Using only a meter stick, test the reflexes of each person
in your group by dropping the meter stick between his
or her thumb and forefinger. One trial only per person,
– We will then identify the top 3 quickest students, and
discuss the results.
• Ball A is dropped from rest at the top of a cliff of
height h as shown. Using g as the acceleration due
to gravity, derive an expression for the time it will
take for the ball to hit the ground.
v = v0 + at
A
v = 0 + gt
h
v = gt
t = v/g
Derive an equation using g, h
y = v0t + 1/2at2
t = √ (2h / g)
h = 0 t + 1/2gt2
• Ball B is projected vertically upward from the foot of
the cliff with an initial speed of vo. Derive an
expression for the maximum height ymax reached by
the ball.
Derive an equation using g, ymax
h
v0
B
v2 = v02 + 2a(∆y)
0 = v02 + 2g(ymax)
ymax = -v02 / (2g)
• Ball A is dropped from rest at the top of the cliff at exactly
the same time Ball B is thrown vertically upward with
speed v0 from the foot of the cliff such that Ball B will
collide with Ball A. Derive an expression for the amount of
time that will elapse before they collide.
yA = v0t + 1/2at2
yA = 0 + 1/2at2
vB = v0B + at
0 = v0B + at
a= - v0B / t
a = 2yA / t2
- v0B / t = 2yA / t2
t2/ t = 2ya /- v0B
t = - 2yA / v0B
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