9/15 do now
• Finish chapter 1 test
Homework:
2.1, 2.3, 2.5, 2.7
No Post Tomorrow
Chapter 2
Motion Along a
Straight Line
Goals for Chapter 2
• To study motion along a straight line
• To define and differentiate average and instantaneous
linear velocity
• To define and differentiate average and instantaneous
linear acceleration
• To explore applications of straight-line motion with
constant acceleration
• To examine freely falling bodies
• To consider straight-line motion with varying acceleration
2.1 Displacement, time, and the average velocity
• Displacement: change in position, it is a vector
quantity. Its direction is from start to end.
∆x = x2 – x1
• Time: change in time
∆t = t2 – t1
• Average x-velocity: the displacement, ∆x,
divided by the time interval ∆t .
vav x
x x2  x1


t t 2  t1
Distance and Average Speed
• Distance: length of the path, it depends on the
path. It is a scalar quantity. It has no direction.
• Average x-speed: the distance traveled ∆s
divided by the time interval ∆t. It is a scalar.
Average speed vs. average velocity
When Alexander Popov set a world record in 1994 by
swimming 100.0 m in 46.74 sec, his average speed was
(100.0 m) / (46.74 s) = 2.139 m/s. but because he swam
four lengths in a 25 meter pool, he started and ended at
the same point and he had zero total displacement and
zero average velocity!
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.
• Note: you can choose any way as +.
P-T graph of the car
X (m)
x2 = 277m
∆x
∆t
x1 = 19m
t (s)
t=1s
t=4s
Check your understanding 2.1
•
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
Practice 2.2
•
In an experiment, a shearwater (a seabird)
was taken from its nest, flown 5150 km away,
and released. The bird found its way back to
its nest 13.5 days after release. If we place the
origin in the nest and extend the +x-axis to the
release point, what was the bird’s average
velocity in m/s
-4.42 m/s
1. For the return flight?
2. For the whole episode, from leaving the nest to
returning?
0 m/s
Practice 2.4
•
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
4.4 m/s
1. Your average speed from pillar to post,
2. You average velocity from pillar to post. -0.72 m/s
Practice 2.6
•
Two runners start simultaneously from 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
Practice 2.8
•
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. Calculate the
average velocity of the car for each time
interval:
a. t = 0 to t = 2.00 s;
b. t = 0 to t = 4.00 s
c. t = 2.00 s to t = 4.00 s.
example
• A cat runs along a straight line (the x-axis) from point A
to point B to point C, as shown. The distance between
points A and C is 5.00 m, the distance between points B
and C is 10.0 m, and the positive direction of the x-axis
points to the right. The time to run from A to B is 20.0 s,
and the time from B to C is 8.00 s.
B
C
A
1. What is the average speed of the cat between points A
and C?
2. What is the average velocity of the cat between points
A and C?
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 < 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.
D/2
t Rick
½ tTim tTim tRick
D2 D2 D 1 1


 (  )
vr
vw
2 vr v w
t
d.
e.
vRick 
D
t Rick
2(vr  vw )

vr  v w
tTim
tTim
D  vr ( )  v w ( )
2
2
2D
tTim 
vr  v w
9/16 do now – on a new sheet
•
Vectors V1 and V2 shown above have equal magnitudes. The vectors
represent the velocities of an object at times t1 and t2, respectively. The
average acceleration of the object between time t1 and t2 was
N
A. Zero
N
B. Directed north
C. Directed west
V2
V1
W
E
W
E
D. Directed north of east
E. Directed north of west
S
Time t1
Homework questions?
Homework: 2. 9, 2.11 and work sheet
S
Time t2
2.2 Instantaneous velocity
• Instantaneous velocity is defined as 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
compute 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.
• In the language of calculus, the limit of ∆x / ∆t as ∆t
approaches zero is called the derivative of x with the
respect to t and is written dx/dt.
• The instantaneous velocity is the limit of the average
velocity as the time interval approaches zero; it
equals the instantaneous rate of change of position with
time.
x dx
v x  lim
t  0
t

dt
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 attack, 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
9/17 do now
• Vectors A and B are shown. Vector C is given by C=B−A.
The magnitude of vector A is 16.0 units, and the
magnitude of vector B is 7.00 units. What is the magnitude
of C?
Homework: 2.9, 2.11, worksheet, 2.55. 2.57,
Example 2.1 Average and instantaneous velocity
Average and instantaneous velocities in x-t graph
Secant line –
average velocity
tangent line –
instantaneous
velocity
example
The automobiles
make a 5 hour trip
over a total
distance of 200 km.
1. Which car starts
later?
2. When does A & B
pass each other?
3. Which car reaches
200 km first?
4. Calculate average speed of A and B.
The Derivative…aka….The
SLOPE!
• Suppose an eccentric pet ant is constrained to move
in one dimension. The graph of his displacement as a
function of time is shown below.
B
x(t +t)
x(t)
A
t
t + t
At time t, the ant is located at
Point A. While there, its
position coordinate is x(t).
At time (t+t), the ant is
located at Point B. While
there, its position coordinate is
x(t + t)
The secant line and the slope
Suppose a secant line is drawn between points A and B.
Note: The slope of the secant line is equal to the rise over
the run.
B
x(t +t)
x(t)
A
t
t + t
The slope of the secant line is average velocity
The “Tangent” line
READ THIS CAREFULLY!
• If we hold POINT A fixed while allowing t to become
very small. Point B approaches Point A and the secant
approaches the TANGENT to the curve at POINT A.
B
B
x(t +t)
x(t)
x(t +t)
A
x(t)
t
t + t
A
t
t + t
We are basically ZOOMING in at point A where upon inspection the line
“APPEARS” straight. Thus the secant line becomes a TANGENT LINE.
The slope of the tangent line is ____________________ velocity.
The derivative
Mathematically, we just found the slope!
y2  y1 x(t  t )  x(t )
slope 

 slope of secant line
x2  x1
t
lim t 0
x(t  t )  x(t )
 slope of tangent line
t
Lim stand for “__________" and it shows the ∆t
approaches zero. As this happens the top numerator
approaches a finite #.
This is what a derivative is. A derivative yields a NEW
function that defines the rate of change of the original function
with respect to one of its variables. The above example
shows the rate of change of "x" with respect to time.
• In most Physics books, the derivative is
written like this:
dx
Mathematicians treat dt as a SINGLE SYMBOL which
means find the derivative. It is simply a mathematical
operation.
The bottom line: The derivative is the
slope of the line tangent to a point on
a curve.
example
• Consider the function x(t) = 3t +2; What is the time rate
of change of the function (velocity)?
• This is actually very easy! The entire equation is linear and
looks like y = mx + b . Thus we know from the beginning
that the slope (the derivative) of this is equal to 3.
We didn't even need
to INVOKE the limit
because the ∆t is
cancel out.
Regardless, we see
that we get a
constant.
Example
• Consider the function x(t) = kt3, where k =
proportionality constant.
dx(t )
x(t  t )  x(t )
 lim t 0
dt
t
[k (t  t )3  k (t )3 ]
 lim t 0
t
t 3  3t 2 (t )  3t (t ) 2  (t )3  t 3
 k  lim t 0
t
 k lim t 0 [3t 2  3t (t ) 2  (t )3 ]
 k (3t 2 )  3kt 2
What happened to
all the ∆t's ? They
went to ZERO when
we invoked the
limit!
What does this all
mean?
The MEANING?
d (kt3 )
2
 3kt
dt
• For example, if t = 2 seconds, using x(t) =
kt3=(1)(2)3= 8 meters.
• The derivative, however, tell us how our
DISPLACEMENT (x) changes as a function of
TIME (t). The rate at which Displacement
changes is also called VELOCITY. Thus if we
use our derivative we can find out how fast the
object is traveling at t = 2 second. Since dx/dt =
3kt2=3(1)(2)2= 12 m/s
THERE IS A PATTERN HERE!!!!
2
d (kt )
1
 2kt
dt
4
d (kt )
3
 4kt
dt
5
d (kt )
4
 5kt
dt
1.Derivative of a constant
d
(C )  0
dx
Example x = 5,
dx
?
dt
Why?
2. Power Rule
d n
n 1
(x )  n  x
dx
Example
x = t5
x=
t-5
x=t
dx
?
dt
dx
?
dt
dx
?
dt
3. Constant Multiplier
d
d
[c  f ( x)]  c  [ f ( x)]
dx
dx
Example
dx
?
dt
x = 4t5
4. Addition and Subtraction Rule
d
d
d
[ f ( x)  g ( x)] 
f ( x) 
g ( x)
dx
dx
dx
The derivative of the sum (or difference) of two or more
functions is the sum (or difference) of the derivatives of the
functions.
Example
x
=2t5
+
3t-1
dx
?
dt
Chain rule
If x is a function of f, and f is a function of t, so
indirectly, x is a function of t: x(f(t))
dx dx df


dt df dt
Example
1 2
x  (2t  3t )
5
f  2t  3t
5
x f
2
1
dx
?
dt
Class work
•
1.
2.
3.
4.
Find the derivatives (dx/dt) of the
following function
x = t3
x = 1/t = t-1
x = (6t3 + 2/t)-2
x = 16t2 – 16t + 4
Average velocity vs. instantaneous
velocity 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
•
If s is distance and t is time, what must
be the dimensions of C1, C2, C3, and C4
in each of the following equations?
a. S = C1t
b. S = ½ C2t2
c. S = C3sinC4t
Hint: the argument of any trigonometric function must be
dimensionless.
Example
• An object moves vertically according to
y(t) = 12 – 4t+ 2t3. what is its velocity at
t = 3 s?
92 m/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
example
•
An object is forced to move along the x axis in
such a way that is displacement is given by x =
30 + 20t – 15t2 where x is in m and t is in s.
a. Find expressions for the velocity of the object.
b. At what time and distance from the origin is the
velocity zero?
t = 0.67 s; x = 36.7 m
c. At what time and location is the velocity -50
m/s
t = 2.3 s; x = - 5.0 m
example
• The position of a vehicle moving on a
straight track along the x-axis is given by
the equation x(t) = 3t3 – 2t2 + t, where x is
in meters and t is in seconds. What is its
velocity at time t = 3s?
70 m/s
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
velocity at time t = 5 s?
13 m/s
example
• The position of a particle moving along the
x-axis is given by the equation x(t) = 2 +
6t2, where x is in meters and t is in
seconds. What is the average velocity
during the interval t = 0 to t = 0.5 s?
3 m/s
example
• An object’s motion is given by the equation
x(t) =2 + 4t3. what is the equation for the
object’s velocity?
v(t) = 12t2
Follow the motion of a particle
– The motion of the particle may be described from x-t
graph.
v
Questions
o
x
t
x
a.
t
o
•
The graph above
shows velocity v
versus time t for an
object in linear
motion. Which of the
following is a
possible graph of
position x versus
time t for this
object?
b. o
x
x
c.
d.
t
o
o
x
e.
o
t
t
t
Test your understanding 2.2
•
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.
P
At which points is vx positive?
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, R, and S from fastest to slowest.
R, P, Q = S
Example 2.10
•
a.
b.
c.
d.
e.
A physics professor leaves her house and walks along the
side walk toward campus. After 5 min it starts to rain and
she returns home. According to the graph, at which of the
labeled points is her velocity
IV
Zero?
Constant and positive?
I
Constant and negative? V
Increasing in magnitude? II
Decreasing in magnitude? III
example
a. In which of the following is the rate of change of the
particle’s momentum zero? I, II
b. In which of the following is the particle’s acceleration
constant?
I, II, III
d
d
d
t
I
t
II
t
III
example
•
In which of these cases is the rate of change of the
particle’s displacement constant?
II
v
v
v
t
I
t
II
t
III
example
• Which pair of graphs represents the same
1-dimensional motion?
A.
B.
C.
D.
example
•
The graph represents the relationship between
distance and time for an object. What is the
instantaneous speed of the object at
0
a. t = 5.0 seconds?
b. t = 2.0 seconds?
1.5 m/s
example
• The graph represents the relationship between the
displacement of an object and its time travel along a
straight line. What is the average speed of the object
during the first 4.0 seconds?
2 m/s
example
d
t
•
a.
b.
c.
d.
e.
o
According to the graph, the velocity of the object must
be
Zero
Constant and positive
Constant and negative
Increasing
decreasing
example
d
t
•
a.
b.
c.
d.
e.
o
According to the graph, the acceleration of the object
must be
Zero
Constant and positive
Constant and negative
Increasing
decreasing
Example
•
a.
b.
c.
x(
15
m)
The graph of an
object’s motion
(along a line) is
shown. Find
the instantaneous
10
velocity of the object
at points A and B. 0.5 m/s
the object’s average
velocity. 0.5 m/s
5
its acceleration.
B
A
0
0
0
10
20
t (s)
Example
•
a.
b.
c.
d.
x(
m)
15
Refer to the graph.
The object’s motion
is represented by the
curve. Find
10
the instantaneous
velocity at point F.
the instantaneous
velocity at point D.
the instantaneous
5
velocity at point C.
the instantaneous
velocity at point E.
0
D
C
F
E
0
10
20
t (s)
Example
A girl walks along an east-west street, and a graph of her
displacement from home is shown in the graph. Find her
a. Average velocity for the whole time interval
0
b. Instantaneous velocity at A, B and C
Distance east
c. Average
velocity for the
(m) B
time interval t = 40
7 min to t = 14
A
min
-10 m/s 20
d. The
instantaneous
velocity at t =
0
5
13.5 min and t
= 15 min
--14 m/s; 6 m/s20
6.7 m/s
C
1
0
1
5
t
2min
0
example
•
1.
2.
A car with an initial positive velocity slows to a stop
with a constant acceleration.
Which graph best represents its position vs. time
graph?
Which graph best represents the velocity vs. time
graph?
A
B
C
t
t
D
t
E
t
t
v
example
• An object moves with a
velocity vs. time graph as
shown. The position vs. time
graph for the same time
period would be
A
x
x
B
t
C
t
x
x
t
t
x
D
E
t
t
example
• An object is moving in a straight line (the x-axis). The graph
shows the x-coordinate of this object as a function of time.
Which one of the following statements about this object is
correct?
a. Between points A and B, both the
x-component of its average velocity
and its average speed are greater
than 0.75 m/s.
b. Between points A and B, both the
x-component of its average velocity
and its average speed are less than
0.75 m/s.
c. Between points A and B, the xcomponent of its average velocity is
0.75 m/s, but its average speed is
greater than 0.75 m/s.
d. Between points A and B, both the
x-component of its average velocity
and its average speed are equal to
0.75 m/s.
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 v  v divided by the
2
1
time interval.
aavg
 

v2  v1 v


t 2  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 2.2
• An astronaut has left an orbiting spacecraft to test a
personal maneuvering unit. As she moves along a
straight line, her partner on the spacecraft measures her
velocity every 2.0 s, starting at time t = 1.0 s:
• Find the average xt
acceleration, and describe
1.0 s
whether the speed of the
astronaut increases or
3.0 s
decreases, for each of these
5.0 s
time intervals:
a. t1 = 1.0 s to t2 = 3.0s
7.0 s
vx
0.8 m/s
t
9.0 s
vx
-0.4 m/s
1.2 m/s 11.0 s -1.0 m/s
1.6 m/s 13.0 s -1.6 m/s
1.2 m/s 15.0 s -0.8 m/s
b. t1 = 5.0 s to t2 = 7.0 s
a. 0.2 m/s2; speed increases
c. t1 = 9.0 s to t2 = 11.0 s
b. -0.2 m/s2; speed decreases
d. t1 = 13.0 s to t2 = 15.0 s
c. -0.3 m/s2; speed increases
d. 0.4 m/s2; speed decreases
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
The instantaneous acceleration is the limit of average
acceleration as the time interval approaches zero.
a  lim t 0

v dv

t dt
dx
v
dt
d dx
d 2x
a ( ) 2
dt dt
dt
Average and instantaneous acceleration
Example 2.3
•
1.
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
Find the change in x-velocity of the car in the time
interval between t1 = 1.0 s and t2 = 3.0 s.
Find the average x-acceleration between t1 = 1.0 s and
t2 = 3.0 s.
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. 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
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 )
Example
•
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
• An object beginning at the origin moves in one
dimension according to v(t) = 12/(6 + 7t).
Determine the acceleration of the object.
example
• The equation of the position of an object moving along
the x-axis is given by x(t) = 1.5t3 – 4.5t2 + .5t, where x is
in meters and t is in seconds. What is the object’s
displacement when its instantaneous acceleration is
equal to zero?
-2.5 m
example
• The velocity of a particle moving along the x-axis
is given by the equation v(t) = 1 + 5t + 2t2, where
v is in m/s and t is in seconds. What is the
average acceleration during the interval t = 0 to t
9 m/s2
= 2?
example
• The position of a particle moving along te x-axis is given
by the equation x(t) = 1 + 2t2 + 3t3, where x is in meters
and t is in seconds. What is the average acceleration
during the interval t = 0 to t = 1?
13 m/s2
example
• An object’s motion is given by the equation x(t) = 4t + 4t3.
what is the equation for the object’s acceleration?
a(t) = 24t
example
• A truck moving along a straight road at 30 m/s applies its
breaks such that its velocity is given by the equation v(t)
= 30 – 2t, where v is in m/s and t is in seconds. What is
the truck’s acceleration at t = 1 s?
-2 m/s2
3.59 (3000)
•
a.
b.
c.
d.
An objects is forced to move along the x axis in such a
way that ist displacement is given by x = 30 + 20t – 15t2
where x is in m and t is in s.
Find expressions for the velocity and acceleration. Is
the acceleration constant? v = (20 – 30t) m/s a =– 30 m/s2
What are the initial position and the initial velocity of the
object?
xo = 30 m; vo = 20 m/s
at what time and distance from the origin is the velocity
zero?
at what time and location is the velocity -50 m/s
t = 2.33s; x = -5.0 m
3.63 (3000)
•
A mass at the end of a spring vibrates up and down
according to the equation y = 8sin(1.5t) cm, where t is
the time in seconds and the complete argument (angle)
of the sine function, 1.5t is in radian.
1. What is the velocity of the mass at t = 0.75 s?
2. At t = 3.0 s?
3. What is the maximum velocity of the mass?
1. 5.2 cm/s
2. -2.5 cm/s
3. v = (+ or -) 12.0 cm/s
Example
•
1.
2.
The velocity of a car traveling on a straight track along
the y-axis is given by the equation v(t) = -12t2 + 6t + 2,
where x is in meters and t is in seconds.
-88 m/s
What is its velocity at time t = 3 s?
What is its acceleration at time t = 3 s?
-66 m/s/s
Finding acceleration on a vx-t graph and ax-t graph
• Average acceleration can be determined by v-t graph
Finding the acceleration on v-t graph
• A graph of  and t may be used to find the acceleration.
• Average acceleration: the slope of secant line.
• Instantaneous acceleration: the slope of a tangent line at point.
Caution: 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
Finding x, v, a in x-t graph
point x
v
A
Neg.
B
0
C
pos.
D
E
pos.
pos.
a
0
0
0
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
Check your understanding 2.3
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.,
position
example
1. At which of the labeled
points is the magnitude of
the velocity greatest?
C
B
A
D
D
E
time
2. At which of the labeled
points is the velocity zero?
C, E
3. At which of the labeled
points is the magnitude of
the acceleration greatest?
E
example
• A child standing on a bridge throws a rock straight down.
The rock leaves the child's hand at t= 0. Which of the
graphs shown here best represents the velocity of the
stone as a function of time?
Av
B
v
v
C
t
t
v
D
E
t
v
t
t
example
x
Which of the following graphs
best describes the xcomponent of the velocity as a
function of time for this object?
t
v
v
A
v
B
C
t
t
v
t
v
D
E
t
t
exercise 2.35
• Two cars, A and B, move along the x-axis. The figure is
a graph of the positions of A and B versus time.
a. At what time(s), if any, do A
and B have the same
position? 1 s, 3 s
b. sketch velocity versus time
for A & B .
v
c. At what time(s), if any, does
A pass B?
3s
d. At what time(s), if any, does
B pass A?
1s
v
t
A
t
B
2.3 motion with constant acceleration
Given:
derive:
v xav
x  x0

t
a xav
vx  vx 0

t
vxav
vx  vx 0

2
(assume t0 = 0)
1.
vx = vx0 + axt
2.
x = x0 + vx0 + ½ axt2
3.
vx2 – vx02 = 2ax(x – x0)
Homework – extra credit
Motion with constant acceleration vx-t graph
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
Kinematics equations for constant
acceleration
Example 2.4
A motorcyclist heading east through a small Iowa city
accelerates after he passes the signpost marking the city
limits. His acceleration is a constant 4.0 m/s2. At time t = 0
he is 5.0 m east of the signpost, moving east at 15 m/s.
a. Find his position and velocity at time t = 2.0 s.
b. Where is the motorcyclist when his velocity is 25 m/s?
Example 2.5
A motorist traveling with a constant speed of 15 m/s passes
a school crossing corner, where the speed limit is 10 m/s.
Just at the motorist passes, a pplice officer on a motorcycle
stopped at the corner starts off in pursuit with constant
acceleration of 3.0 m/s2.
a. How much time elapses before the officer catches up
with the motorist?
b. What is the officer’s speed at that point?
c. What is the total distance each vehicle has traveled at
that point?
Test your understanding 2.4
Four possible vx-t graphs are shown for the two vehicles
in example 2.5. which graph is correct?
example
1. At what time after t = 0 does the object again pass through
its initial position?
9s
2. During which interval does the particle have the same
average acceleration as 12 s < t < 14 s?
v (m/s)
a. 9s < t < 11s
b. 2s < t < 5s
20
10
c. 0s < t < 3s
d. 3s < t < 7s
e. 5s < t < 11s
0
5
-10
-20
10
14
Example 2.22
The catapult of the aircraft carrier USS Abraham Lincoln accelerates an F/A18 Hornet jet fighter from rest to a takeoff speed of 173 mi/h in a distance of
307 ft. Assume constant acceleration.
a. Calculate the acceleration of the fighter in m/s2
b. Calculate the time required for the fighter to accelerate to takeoff speed.
• 1 mi/hr = 0.447 m/s
• 1 ft = 0.3048 m
173 mi/h = 77.3 m/s
307 ft = 93.6 m
a.
ax = 32.0 m/s2
b.
t = 2.42 s
v
Example
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
2.5 Free Falling Bodies
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 = 9.81 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
Example 2.6
A one-euro coin is
dropped from the
Leaning Tower of
Pisa. It starts from
rest and falls freely.
Compute its position
and velocity after .1.0
s. 2.0 s, and 3.0 s.
Example 2.7
You throw a ball vertically upward
from the roof of a tall building. The
ball leaves your hand at a point even
with the roof railing with an upward
speed of 15.0 m/s; the ball is then in
free fall. On its way back down, it just
misses the railing. At the location of
the building, g = 9.80 m/s2. find
a. The position and velocity of the
ball 1.00 s and 4.00 s after leaving
your hand
b. The velocity when the ball is 5.00
m above the railing
c. The maximum height reached and
the time at which it is reached
d. The acceleration of the ball when it
is at its maximum height.
Velocity and acceleration at the highest point
Example 2.8
Find the time when the ball in Example 2.7 is 5.00 m below
the roof railing.
Check your understanding 2.5
•
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
Example
• 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)
Example
• 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
Example
• 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
2.6 velocity and position by integration
Finding v(t) and x(t) when given a(t)
• In the case of straight-line motion, if the position x is a
known function of time, we can find vx = dx/dt to find xvelocity. And we can use ax = dvx/dt to find the xacceleration as a function of time
• In many situations, we can also find the position and
velocity as function of time if we are given function ax(t).
The “AREA”
In v-t graph, the area under the line
represent displacement.
v (m/s)
v (m/s)
x = area
x = area
t(s)
t1
t2
How ever, if
acceleration is not
constant, how can we
determine x(t)?
t(s)
t1
t2
v (m/s)
v(t)
t
t
t(s)
Zoom in
We have learned that the rate of
change of displacement is defined as
the VELOCITY of an object. Consider
the graph below
dx  v(t )  dt
v (m/s)
v  lim t 0
x
t
dx
v
dt
∆Area = v(t)∙dt
TOTAL DISPLACEMENT = ∑∆Area = ∑v(t)∙dt
v(t)
dt
t1
t
t2
t(s)
The “Integral” – the area
• The temptation is to use the conventional summation
sign “S" . The problem is that you can only use the
summation sign to denote the summing of DISCRETE
QUANTITIES and NOT for something that is
continuously varying. Thus, we cannot use it.
• When a continuous function is summed, a different sign
is used. It is called an Integral, and the symbol looks like
this:
When you are dealing with a situation where you have to integrate, realize:
• WE ARE GIVEN: the derivative already
• WE WANT: The original function x(t)
So what are we basically doing?
WE ARE WORKING BACKWARDS!!!!! OR FINDING THE ANTI DERIVATIVE
Example
• An object is moving at
velocity with respect to
time according to the
equation v(t) = 2t.
a) What is the displacement
function? Hint: What was
the ORIGINAL
FUCNTION BEFORE the
“derivative? was taken?
b) How FAR did it travel from
t = 2s to t = 7s?
x(t )   v dt   (2t )dt 
x(t )  t 2
These are your LIMITS!
t 7
t 7
t 2
t 2
x(t )   v dt   (2t ) dt tt27 t 2
7 2  2 2  49  4  45 m
You might have noticed that in the above example we had to find the
change() over the integral to find the area, that is why we subtract. This
might sound confusing. But integration does mean SUM. What we are doing
is finding the TOTAL AREA from 0-7 and then the TOTAL AREA from 0-2.
Then we can subtract the two numbers to get JUST THE AREA from 2-7.
In summary…
• So basically derivatives are
used to find SLOPES and
Integrals are used to find
AREAS.
dx
v
dt
dv
a
dt
x   v dt v   a dt
Example
Here is a simple example of which you
may be familiar with:
Assume we know the circumference
of a circle is 2pr, where r is the
radius. How can we derive an
expression for the area of a circle
We begin by taking a differential
whose
radius is R?
HOOP of radius "r" and
differential
thickness “dr” as shown.
If we determine the area of JUST OUR CHOSEN HOOP, we could do
the calculation for ALL the possible hoops inside the circle.
Having done so, we would then SUM up all of those hoops to find the
TOTAL AREA of the circle. The limits are going to be the two extremes,
when r = R and when r = 0
Backward power rule
d
n+1) = (n+1) xn
(x
dx
xn+1
∫xn dx = n+1 + C
bn+1
an+1
∫axndx = n+1 - n+1
General expression, no
limits
Evaluation with given
limits
b
Special case: n = 0
b
∫ dx = x + C
or
∫ a dx = b - a
Addition/subtraction rule
constant multiplier rule
∫ (u +
v) dx = ∫ u dx + ∫ v dx
∫ au dx = a ∫ u dx
Example 2.9
• Sally is driving along a straight highway in her classic 1965
Mustang. At time t = 0, when Sally is moving at 10 m/s in the
positive x-direction, she passes a signpost at x = 50 m. her xacceleration is a function of time:
a x  2.0 m / s 2  (0.10 m / s 3 )t
a.
b.
c.
d.
find her x-velocity and position as functions of time
When is her x-velocity greatest?
What is the maximum x-velocity?
Where is the car when it reaches the maximum x-velocity?
Position, velocity and acceleration of the car
Example 2.10
Use Eqs.
t
v x  v0 x   a x dt
0
t
x  xo   v x dt
0
To find vx and x as functions of time in the case
in which the acceleration is constant.
Example
A particle moving in one dimension has a position
function defined as: x(t) = 6t4-2t
a) At what point in time does the particle change its
direction along the x-axis?
b) In what direction is the body traveling when its
acceleration is 12 m/s/s?
dx d (6t 4  2t )
a) v 

0
dt
dt
dv d (24t 3  2)
b) a 

 12
dt
dt
t = 0.437 s
t = 0.408 s v = -0.37 m/s
Example 2.50
The acceleration of a bus is given by ax(t)=αt, where α = 1.2
m/s3.
a. If the bus’s velocity at time t = 1.0 s is 5.0 m/s, what is its
velocity at time t = 2.0 s?
b. If the bus’s position at time t = 1.0 s is 6.0 m, what is its
position at time t = 2.0 s?
c. Sketch ax-t, vx-t and x-t graphs for the motion.
example
• An object initially at rest experiences a time-varying
acceleration given by a = (2 m/s3)t for t >= 0. How far
does the object travel in the first 3 seconds? 9 m
a = (2 m/s3)t
vx = (1 m/s3)t2
vx = vox + (2 m/s3)½ t2
x = xo + (1 m/s3)(1/3)t3
vox = 0
x(3 s) – x (0 s) = (1 m/s3)(1/3)(3 s)3
vx = (1 m/s3)t2
x(3 s) – x (0 s) = 9 m
Check your understanding 2.6
•
If the x-acceleration ax is increasing with
time, will the vx-t graph be
1. A straight line,
2. Concave up
3. Concave down