# Document

```Homework – due Friday, 9/23
Questions: 2, 7, 9, 12, 13, 22, 23, 24, 25, 29,
31, 34, 40, 44 – the solutions are on the
school website.
Homework – due Tuesday, 9/20 –
11:00 pm
Mastering physics wk 3
Chapter 2
Motion along a straight line
•Using mathematics to describe
motion in terms of position, velocity
and acceleration.
•Use concepts of ideal particle, only
consider translational motion
Do now
• At time t = t1, and object’s velocity is given by the vector
v1 a short time later, at t = t2, the object’s velocity is the
vector v2. If the magnitude of v1 = the magnitude of v2,
which one of the following vectors best illustrates the
object’s average acceleration between t = t1 and t = t2
v1
-v1
v2
v2 -v1
A
B
C
D
v2
E
objective
Understand the relationship between
Displacement, Time, and Average Velocity
Distance, Time, and Average Speed
• Distance: length of the path, it depends on the path. It is
a scalar quantity. It has no direction.
• Distance ≥ 0
• speed: how fast the distance is covered. It has no
direction.
• Speed ≥ 0
• Average x-speed: the distance traveled ∆s divided by
the time interval ∆t .
vav-x =
∆s
∆t
Displacement, Time, and Average
Velocity
• Displacement: change in position, it is a vector quantity.
Its direction is from start to end.
∆x = x2 – x1
∆x Displacement can be zero, positive or negative.
-positive if you move to the positive direction
-negative if you move to the negative direction.
• velocity: the rate of change in position, it is a vector
quantity. It has the same direction as its motion.
– Velocity is positive if ∆x is positive
– Velocity is negative is ∆x is negative.
• Average x-velocity: the displacement, ∆x, divided by
∆x
x2 – x1
the time interval ∆t .
vav-x =
=
∆t
t2 – t1
Average speed vs. average velocity
• Average speed is not the magnitude of average velocity.
Let's suppose Taylor covered each leg
of her journey in one second. This
means that the total time for her trip
was 6 seconds.
Average speed = 12 m/ 6s = 2 m/s
Average velocity = 3 m / 6 s = 0.5 m/s
Direction of velocity is east
example
Position at t2 = 4.0 s
Position at t1 = 1.0 s
+displacement
x2 = 277 m
x1 = 19 m
• What is the average velocity of the car?
• Vav-x = (277 m – 19 m) / (4.0 s – 1.0 s) = 86 m/s
• The average velocity is positive because it is
moving in the positive direction.
P-T graph of the car
X (m)
x2 = 277m
∆x
∆t
x1 = 19m
t (s)
t=1s
t=4s
For a displacement along the x-axis, an object’s
average x-velocity vav-x equals the slope of a line
connecting the corresponding points on a graph of
position (x) versus time (t)
•
1.
2.
3.
4.
5.
a.
b.
c.
Each of the following automobile trips takes one hour. The
positive x-direction is to the east.
A travels 50 km due east.
B travels 50 km due west
C travels 60 km due east, then turns around and travels 10
km due west
D travels 70 km due east.
E travels 20 km due west, then turns around and travels 20
km due east.
Rank the five trips in order of average x-velocity from most
positive to most negative. 4, 1, 3, 5, 2
Which trips, if any, have the same average x-velocity? 1, 3
For which trip, if any, is the average x-velocity equal to
zero? 5
example
•
Starting from a pillar, you run 200 m east (the
+x-axis) at an average speed of 5.0 m/s, and
then run 280 m west at an average speed of
4.0 m/s to a post. Calculate
1. Your average speed from pillar to post, 4.4 m/s
2. You average velocity from pillar to post. -0.72 m/s
example
•
Two runners start simultaneously form the same point
on a circular 200 m track and run in the same
direction. One runs at a constant speed of 6.20 m/s,
and the other runs at a constant speed of 5.50 m/s.
1.
When will the fast one first “lap” the slower one and
how far from the starting point will each have run?
286 s, 1770 m, 1570 m
2.
When will the fast one overtake the slower one for the
second time, and how far from the starting point will
they be at that instant?
572 s, 3540 m, 3140 m
Example - Walking 1/2 the time vs.
Walking 1/2 the distance
• Tim and Rick both can run at speed vr and walk at speed
vw, with vw &lt; vr. They set off together on a journey of
distance D. Rick walks half of the distance and runs the
second half. Tim walks half of the time and runs the other
half.
a) Draw a graph showing the positions of both Tim and Rick
versus time.
b) Write two sentences explaining who wins and why.
c) How long does it take Rick to cover the distance D?
d) Find Rick's average speed for covering the distance D.
e) How long does it take Tim to cover the distance?
solution
a.
b. Tim wins because he takes
short time to cover the same
distance as Rick.
x
D
c. tRick = (D ∕ 2)/vr + (D/2)/ vw
D/2
&frac12; tTim tTim tRick
t
tRick = D/2vr + D/2vw
d. vRick = D / tRick
vRick = 2(vr∙vw) ∕ (vw + vr )
e. vr∙(tTim/2) + vw∙(tTim/2) = D
tTim= 2D/(vr + vw)
example
1. Which car starts
later?
2. When does A &amp; B
pass each other?
3. Which car reaches
200 km first?
4. Calculate average speed of A and B.
practice
objective
• Instantaneous velocity
• Determine velocity using x-t graph.
Instantaneous velocity
• Instantaneous velocity: the velocity at any
specific instant of time or specific point
along the path.
• Instantaneous velocity is a vector quantity,
its magnitude is the speed, its direction is
the same as its motion’s direction.
• How long is an instant?
– In physics, an instant refers to a single
value of time.
P2
P1
• To find the instantaneous velocity at point P1, we move
the second point P2 closer and closer to the first point P1
and computer the average velocity vav-x = ∆x / ∆t over the
ever shorter displacement and time interval. Both ∆x and
∆t become very small, but their ratio does not necessarily
become small.
∆x
• In the language of calculus, the limit of ∆t as ∆t
approaches zero is called the derivative of x with the
respect to t and is written as dx
dt
vx = lim ∆x = dx
∆t 0 ∆t
dt
When ∆x is positive, vx is positive
When ∆x is negative, vx is negative
Example 2.1
•
a.
b.
c.
d.
A cheetah is crouched 20 m to the east of an observer’s
vehicle. At time t = 0 the cheetah charges an antelope and
begins to run along a straight line. During the first 2.0 s of
the attach, the cheetah’s coordinate x varies with time
according to the equation x = 20 m + (5.0 m/s2)t2.
Find the displacement of the cheetah between t1 = 1.0 s
and t2 = 2.0 s
Find the average velocity during the same time interval.
Find the instantaneous velocity at time t1 = 1.0 s by taking
∆t = 0.1 s, then ∆t = 0.01 s, then ∆t = 0.001 s.
Derived a general expression for the instantaneous velocity
as a function of time, and from it find vx at t = 1.0 s and t =
2.0 s
Common derivative equations
d xn = n xn-1
dx
d sinx = cosx
dx
d x
e = ex
dx
d Cu(x) = C du
dx
dx
y = g(u); u = f(x)
d C = 0
dx
d cosx = -sinx
dx
d
1
lnx =
dx
x
d (u+v) = du + dv
dx
dx
dx
dy
dy
du
dx = du • dx
Derivatives Practice
•
1.
2.
3.
4.
Find the derivatives (dx/dt) of the
following function
x = t3
x = 1/t
x = 6t3 + 2/t
x = 16t2 – 16t + 4
1.
2.
Example
•
A Honda Civic travels in a straight line
along a road. Its distance x from a stop
sign is given as a function of time t by the
equation x(t) = αt2 – βt3, where α = 1.50
m/s2 and β = 0.0500 m/s3.
1. Calculate the average velocity of the car
for the time interval: t = 0 to t = 4.00 s;
2. Determine the instantaneous velocity of
the car at t = 2.00 s and t = 4.00 s.
example
• An object is moving in one dimension
according to the formula x(t) = 2t3 – t2 – 4.
find its velocity at t = 2 s.
example
• The position of an object moving in a
straight line is given by x = (7 + 10t – 6t2)
m, where t is in seconds. What is the
object’s velocity at 4 seconds?
example
• An object moves vertically according to
y(t) = 12 – 4t+ 2t3. what is its velocity at
t = 3 s?
example
•
a.
b.
c.
d.
e.
An object moves in one dimension such
that x(t) is proportional to t5/2. this means
v2 will be proportional to
t3/2
t7/2
t7
t
t3
velocity on P-T graph
Slope of the line P1P2
represents the average
velocity v between t1 and
t 2.
To get instantaneous
velocity at P1, we pick a
point Pi which is
extremely close to P1:
Instantaneous velocity (velocity) as the limit as we let ∆t →0. it is
equal to the slope of the tangent to the curve at the point.
example
x
C
D
0
A
B
E
t
1. At which point does the
particle has greatest positive
velocity?
B
2. At which point does the
particle has zero velocity? C
3. At which point does the
particle has smallest
nonzero negative velocity?
The slope of the tangent at
E
any point equals the velocity
at that point.
•
a.
b.
c.
d.
e.
According to the graph
Rank the values of the particle’s x-velocity vx at the
points P, Q, R, and S from most positive to most
negative.
At which points is vx positive? P
At which points is vx negative? R
At which points is vx zero? Q, S
Rank the values of the particle’s speed at the points P,
Q
Q, R, and S from fastest to slowest.
P
R, P, Q = S
R
S
Given x = 2.1t2 + 2.80, graph x vs. t and v vs. t
Accelerated motion
Given x = 2.1t + 2.80, graph x vs. t and v vs. t
x
Constant (uniform) motion
Average velocity = instantaneous velocity
2.80
v
t
2.80
t
practice
• Hand out – wizard test maker
objective
• average and instantaneous acceleration
2.3 average and instantaneous
acceleration
• The average acceleration of the particle as it moves
from P1 to P2 is a vector quantity, whose magnitude
equals to the change in velocity (v2 – v1) divided by the
time interval.
aav-x =
v2 – v1
∆v
=
t2 – t1
∆t
Velocity describes how fast a body’s position change with
time.
Acceleration describes how fast a body’s velocity change,
it tells how speed and direction of motion are changing.
example
• A racquetball strikes a wall with a speed of
30 m/s. the collision takes 0.14 s. If the
average acceleration of the ball during
collision is 2800 m/s/s. what is the
rebound speed?
Instantaneous acceleration
aav-x =
v2 – v1
t2 – t1
=
∆v
∆t
• the instantaneous acceleration is the limit of average
acceleration as the time interval approaches zero.
∆v
aav-x = lim
t
0 ∆t
dv
aav-x =
dt
dx
Since v =
dt
2x
d
aav-x =
dt2
dv
dt
d2x
= 2
dt
Example 2.3
•
Suppose the x-velocity vx of a car at any time t is given
by the equation: vx = 60 m/s + (.50 m/s2)t2
1. Find the change in x-velocity of the car in the time
interval between t1 = 1.0 s and t2 = 3.0 s.
2. Find the average x-acceleration between t1 = 1.0 s and
t2 = 3.0 s.
3. Find the average x-acceleration at time t1 = 1.0 s by
taking ∆t to be first 0.1 s, then 0.01 s, then 0.001 s.
4. Derive an expression for the instantaneous xacceleration at any time, and use it to find the xacceleration at t= 1.0 s and t = 3.0 s.
1. 4.0 m/s
2. 2.0 m/s2
3. 1.05 m/s2; 1.005 m/s2; 1.005 m/s2
4. a = (1.0 m/s3)t; 1.0 m/s2; 3.0 m/s2
example
• The position of an object as a function of time is given by
x(t) = at3 – bt2 + ct - d,
where a = 3.6 m/s3, b = 5.0 m/s2; c = 6 m/s; and d = 7.0 m
(a) Find the instantaneous acceleration at t = 2.4 s.
(b) Find the average acceleration over the first 2.4
seconds.
a.42 m/s2
b.16 m/s2
3.60 (3000)
•
a.
b.
A particle moving along the x-axis has a velocity given
by v = 4t – 2.50t2 cm/s for t in seconds. Find its
acceleration at
t = 0.50 s
t = 3.0 s
a. 1.50 cm/s2
b. -11.0 cm/s2
example
• An object moves vertically according to
y(t) = 12 – 4t + 2t3. What is its acceleration
at t = 3 s?
36 m/s2
example
• The position of a vehicle moving on a
straight track along the x-axis is given by
the equation x(t) = t2 + 3t + 5 where x is in
meters and t is in seconds. What is its
acceleration at time t = 5 s? 2 m/s2
Finding acceleration on a vx-t graph
• We can interpret the average and the
instantaneous velocity in terms of the slope of a
graph of position versus time.
• In the same way, we can interpret average and
instantaneous x-acceleration by using vx-t graph.
• The average acceleration aav-x = ∆ v / ∆t during
this interval equals to the slope of the line
between points (t1,v1) and (t2, v2).
• The instantaneous acceleration at point p
equals to the tangent of the curve at point p.
Caution: the signs of x-acceleration and
x-velocity
• How can you determine if object is speeding
up or slowing down?
– If the object’s velocity and acceleration have the
same sign, then the object is speeding up
– If they have opposite sign, then the object is
slowing down
The sign of acceleration and
velocity
a is in the same
direction as v
a is in the opposite
direction as v
v: pos
a: pos.
v: pos
a: neg.
v: neg.
a: neg.
v: neg.
a: pos.
We can obtain an object’s position, velocity and
acceleration from it v-t graph
point
x
A
Given
0
B
Neg.
C
Pos.
D
Pos.
E
Pos.
v
a
0
0
0
Finding acceleration on a x-t graph
On a x-t graph, the acceleration is given by the
curvature of the graph.
Curves up from the point: acceleration is
positive
straight or not curves up or down:
acceleration is zero
Curves down: acceleration is negative
point x
v
A
Neg.
B
0
C
pos.
D
E
pos.
pos.
a
0
0
0
Refer to the graph,
1. At which of the points P, Q, R, and S is the x-acceleration ax
positive?
S
2. At which points is the x-acceleration ax negative? Q
3. At which points does the x-acceleration appear to be zero?
P, R
4. At each point state whether the speed is increasing,
decreasing, or not changing.
P: v is not change;
Q: v is zero, changing from pos. to neg., first
decrease in pos. then increase in neg.,
R: v is neg., constant;
S: v is zero, changing from neg. to pos., first
decrease in neg. then increase in pos.,
Example
• The figure is
graph of the
coordinate of a
spider crawling
along the x-axis.
Graph its velocity
and acceleration
as function of
time.
V
t
a
t
Derive equations for motion with
constant acceleration
Given:
vxav = x – x0
t
axav = vx – vx0
t
derive:
vxav = vx + vxo
2
(assume t0 = 0)
1.
vx = vxo + axt
2.
x = xo + vxo + &frac12; axt2
3.
vx2 – vxo2 = 2ax(x – x0)
Objectives: Motion with constant velocity
Graphs of motion
x
ax
t
t
vx
t
a-t graph
A horizontal line indicate the slope = 0, a = 0
Since ax = ∆v / ∆t; ∆v = ax ∙ ∆t which is
represented by the area.
The area indicate the change in velocity
during ∆t
v-t graph
Slope: indicate
acceleration
Area indicate the
change in
displacement
from time = 0 to t
Kinematics equations for constant
acceleration
Example 2.4
a.
b.
X = 55 m
Example: 2.5
a. t = 10 s.
b. v = 30 m/s.
c. d = 150 m
Four possible vx-t graphs are shown for the two vehicles
in example 2.5. which graph is correct?
An example of constant acceleration –
acceleration due to gravity
If we ignore air friction and the
effects due to the earth’s rotation, all
objects fall and rise at the constant
acceleration.
The constant acceleration of a freely
falling body is called the acceleration due
to gravity, and we use letter g to
represent its magnitude. Near the earth’s
surface g = 9.8 m/s/s = 32 ft/s/s
On the surface of the moon, g = 1.6 m/s/s
On the surface of the sun, g = 270 m/s/s
a = -g
t = 1.0 s
x = -4.9 m; v = -9.8 m/s
t = 2.0 s
x = -19.6 m; v = -19.6 m/s
t = 3.0 s
x = -44.1 m; v = -29.4 m/s
A
B
C
• a = -g = -9.80 m/s/s
-5.00 m = 0 + 15 m/s)(t) – &frac12; (9.80 m/s/s)t2
(9.80 m/s/s)t2 – (15 m/s)(t) – (5.00 m) = 0
•
If you toss a ball upward with a certain initial
speed, it falls freely and reaches a maximum
height h at time t after it leaves your hand.
1. If you throw the ball upward with double the
initial speed what new maximum height does
4h
the ball reach?
2. If you throw the ball upward with double the
initial speed, how long does it take to reach its
maximum height?
2t
Class work
9/28 do now – to be collected
• A uniform cylinder, initially at rest on a frictionless,
horizontal surface, is pulled by a constant force F from
time t = 0 to time t = T. From time t = T on, this force is
removed. Sketch a graph best illustrates the speed, v, of
the cylinder’s center of mass from t = 0 to t = 2T?
v
T
2T
time
v
9/30 Do now
t
The graph above shows velocity v versus time t for an object in linear
motion. Sketch a graph of position x versus time t for this object?
x
t
10/1 do now
• A rock is dropped off a cliff and falls the first half
of the distance to the ground in t1 seconds. If it
falls the second half of the distance in t2
seconds, what is the value of t2/t1? (ignore air
resistance) √2 - 1
10/4 do now
• An object is dropped from rest from the top of a 400 m
cliff on Earth. If air resistance is negligible, what is the
distance the object travels during the first 6 s of its fall?
176 m
10/5 do now
• The position of an object is given by the equating x =
3.0t2 + 1.5 t + 4.5, where x is in meters and t is in
seconds. What is the instantaneous acceleration of the
object at t = 3.00 s?
2
6 m/s
```