9/15 do now finish chapter 1 test homework: 2.1, 2.3, 2.5, 2.7 no post tomorrow

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

• Average x-velocity: the displacement, ∆x, divided by the time interval ∆t .

• Time: change in time ∆t = t2 – t1

12

12

tt

xx

t

xv xav

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

• What is the average velocity of the car?

Position at t1 = 1.0 sPosition at t2 = 4.0 s

x1 = 19 mx2 = 277 m

• 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 +.

+displacement

P-T graph of the car

t (s)

X (m)

x1 = 19m

x2 = 277m

Slope

= v av

-x

∆x

∆t

t = 1 s t = 4 s

Check your understanding 2.1• Each of the following automobile trips takes one hour. The

positive x-direction is to the east.

1. A travels 50 km due east.

2. B travels 50 km due west

3. C travels 60 km due east, then turns around and travels 10 km due west

4. D travels 70 km due east.

5. E travels 20 km due west, then turns around and travels 20 km due east.

a. Rank the five trips in order of average x-velocity from most positive to most negative.

b. Which trips, if any, have the same average x-velocity?

c. For which trip, if any, is the average x-velocity equal to zero?

4, 1, 3, 5, 2

1, 3

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

1. For the return flight?2. For the whole episode, from leaving the nest to

returning?

-4.42 m/s

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

1. Your average speed from pillar to post,

2. You average velocity from pillar to post.

4.4 m/s

-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?

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?

286 s, 1770 m, 1570 m

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 sc. 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?

t

x

D

D/2

tTim½ tTim tRick

b. Tim wins because he takes short time to cover the same distance as Rick.

a.

solution

)11

(2

22

wrwrRick vv

D

v

D

v

Dt

d.

wr

wr

RickRick vv

vv

t

Dv

)(2

c.

wrTim

Timw

Timr

vv

Dt

tv

tvD

2

)2

()2

(e.

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

W E

S

N

W E

S

V1V2

Time t1 Time t2

A. Zero

B. Directed north

C. Directed west

D. Directed north of east

E. Directed north of west

Homework questions?Homework: 2. 9, 2.11 and work sheet

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.

• 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.

P1P2

• 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.

dt

dx

t

xv

tx

0

lim

• 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.

a. Find the displacement of the cheetah between t1 = 1.0 s and t2 = 2.0 s

b. Find the average velocity during the same time interval.

c. 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.

d. 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

Example 2.1

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

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 automobiles make a 5 hour trip over a total distance of 200 km.

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.

t

x(t)

t + t

x(t +t)

A

B

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 isx(t + t)

The secant line and the slopeSuppose 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.

t

x(t)

t + t

x(t +t)

A

B

The slope of the secant line is average velocity

The “Tangent” lineREAD 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.

t

x(t)

t + t

x(t +t)

A

B

t

x(t)

t + t

x(t +t)

A

B

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 derivativeMathematically, we just found the slope!

line tangent of slope)()(

lim

linesecant of slope)()(

0

12

12

t

txttx

t

txttx

xx

yyslope

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:

Mathematicians treat 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.

dt

dx

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.

What happened to all the ∆t's ? They went to ZERO when we invoked the limit!

What does this all mean?22

3220

33223

0

33

0

0

3)3(

])()(33[lim

)()(3)(3lim

])()([lim

)()(lim

)(

kttk

ttttk

t

tttttttk

t

tkttk

t

txttx

dt

tdx

t

t

t

t

The MEANING?

• 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

23

3)(

ktdt

ktd

THERE IS A PATTERN HERE!!!!

45

34

12

5)(

4)(

2)(

ktdt

ktd

ktdt

ktd

ktdt

ktd

Example x = 5,

1.Derivative of a constant

0)( Cdx

dWhy?

?dt

dx

2. Power Rule

1)( nn xnxdx

d

x = t5

Example

x = t-5

x = t

?dt

dx

?dt

dx

?dt

dx

3. Constant Multiplier

)]([)]([ xfdx

dcxfc

dx

d

Example

?dt

dx

x = 4t5

4. Addition and Subtraction Rule

)()()]()([ xgdx

dxf

dx

dxgxf

dx

d

The derivative of the sum (or difference) of two or more functions is the sum (or difference) of the derivatives of the functions.

x =2t5 + 3t-1

Example

?dt

dx

Chain ruleIf x is a function of f, and f is a function of t, so indirectly, x is a function of t: x(f(t))

Example

dt

df

df

dx

dt

dx

?dt

dx215 )32( ttx

2

15 32

fx

ttf

Class work• Find the derivatives (dx/dt) of the

following function

1. x = t3

2. x = 1/t = t-1

3. x = (6t3 + 2/t)-2

4. 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

• An object moves in one dimension such that x(t) is proportional to t5/2. this means v2 will be proportional to

a. t3/2

b. t7/2

c. t7

d. t

e. 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?

c. At what time and location is the velocity -50 m/s

t = 0.67 s; x = 36.7 m

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.

Questions

• 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?

v

tox

to

x

to

x

to

x

to

x

to

x

t

x

to

a. b.

c. d.

e.

Test your understanding 2.2• According to the graph

a. Rank the values of the particle’s x-velocity vx at the points P, Q, R, and S from most positive to most negative.

b. At which points is vx positive?

c. At which points is vx negative?

d. At which points is vx zero?

e. Rank the values of the particle’s speed at the points P, Q, R, and S from fastest to slowest.

P

R

Q, S

R, P, Q = S

Example 2.10• 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

a. Zero?

b. Constant and positive?

c. Constant and negative?

d. Increasing in magnitude?

e. Decreasing in magnitude?

IV

I

VII

III

examplea. In which of the following is the rate of change of the

particle’s momentum zero?

b. In which of the following is the particle’s acceleration constant?

t

d

t

d

t

d

I II III

I, II

I, II, III

example

• In which of these cases is the rate of change of the particle’s displacement constant?

t

v

t

v

t

v

I II III

II

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

a. t = 5.0 seconds?

b. t = 2.0 seconds?

0

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

• According to the graph, the velocity of the object must be

a. Zerob. Constant and positivec. Constant and negatived. Increasinge. decreasing

t

d

o

example

• According to the graph, the acceleration of the object must be

a. Zerob. Constant and positivec. Constant and negatived. Increasinge. decreasing

t

d

o

Example• The graph of an

object’s motion (along a line) is shown. Find

a. the instantaneous velocity of the object at points A and B.

b. the object’s average velocity.

c. its acceleration.

t (s)0

0 10 20

x(m)

5

10

15

A

B

0.5 m/s

0.5 m/s

0

• Refer to the graph. The object’s motion is represented by the curve. Find

a. the instantaneous velocity at point F.

b. the instantaneous velocity at point D.

c. the instantaneous velocity at point C.

d. the instantaneous velocity at point E.

t (s)00 10 20

x(m)

5

10

Example15

C

D

F

E

ExampleA 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

b.Instantaneous velocity at A, B and C

c. Average velocity for the time interval t = 7 min to t = 14 min

d. The instantaneous velocity at t = 13.5 min and t = 15 min

0

40

Distance east (m)

20

-20

5 10

15

20

t min

A

B

C

0

6.7 m/s

-10 m/s

--14 m/s; 6 m/s

example• A car with an initial positive velocity slows to a stop

with a constant acceleration.

1. Which graph best represents its position vs. time graph?

2. Which graph best represents the velocity vs. time graph?

t t t

t t

A B C

ED

example• An object moves with a

velocity vs. time graph as shown. The position vs. time graph for the same time period would be

t

v

t t t

t

x x x

x

t

x

A B C

ED

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 x-component 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 divided by the time interval.

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.

12 vv

t

v

tt

vvaavg

12

12

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:

t vx t vx

1.0 s 0.8 m/s 9.0 s -0.4 m/s

3.0 s 1.2 m/s 11.0 s -1.0 m/s

5.0 s 1.6 m/s 13.0 s -1.6 m/s

7.0 s 1.2 m/s 15.0 s -0.8 m/s

• Find the average x-acceleration, and describe whether the speed of the astronaut increases or decreases, for each of these time intervals:

a. t1 = 1.0 s to t2 = 3.0s

b. t1 = 5.0 s to t2 = 7.0 s

c. t1 = 9.0 s to t2 = 11.0 s

d. t1 = 13.0 s to t2 = 15.0 s

a. 0.2 m/s2; speed increases

b. -0.2 m/s2; speed decreases

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

dt

dv

t

va t

0lim

2

2

)(dt

xd

dt

dx

dt

da

dt

dxv

The instantaneous acceleration is the limit of average acceleration as the time interval approaches zero.

Average and instantaneous acceleration 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. Derive an expression for the instantaneous x-acceleration at any time, and use it to find the x-acceleration 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 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

a. t = 0.50 s

b. 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 = 2? 9 m/s2

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)• 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.

a. Find expressions for the velocity and acceleration. Is the acceleration constant?

b. What are the initial position and the initial velocity of the object?

c. at what time and distance from the origin is the velocity zero?

d. at what time and location is the velocity -50 m/s

v = (20 – 30t) m/s

xo = 30 m; vo = 20 m/s

a =– 30 m/s2

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

• 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.

1. What is its velocity at time t = 3 s?

2. What is its acceleration at time t = 3 s?

-88 m/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

v: pos

a: pos.

v: neg.

a: neg.

a is in the opposite direction as v

v: pos

a: neg.

v: neg.

a: pos.

We can obtain an object’s position, velocity and acceleration from it v-t graph

point x v a

A Given 0

B

C

D

E

Neg.

Pos.

Pos.

Pos.

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

A

B

C

D

E

Neg.

0

pos.

pos.

pos.

0

0

0

Finding x, v, a in x-t graph

Example

V

t

t

a

• The figure is graph of the coordinate of a spider crawling along the x-axis. Graph its velocity and acceleration as function of time.

Check your understanding 2.3Refer to the graph, 1. At which of the points P, Q, R, and S is the x-acceleration

ax positive?

2. At which points is the x-acceleration ax negative?

3. At which points does the x-acceleration appear to be zero?

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.,

S

Q

P, R

examplep

ositi

on

time

A

BC

D

E

1. At which of the labeled points is the magnitude of

the velocity greatest?

2. At which of the labeled points is the velocity zero?

3. At which of the labeled points is the magnitude of the acceleration greatest?

D

C, E

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?

t

v

t

v

t

v

t

v

t

v

A B C

D E

examplex

t

Which of the following graphs best describes the x-

component of the velocity as a function of time for this object?

t

v

t

v

t

v

t

v

t

v

A B C

D E

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?

b. sketch velocity versus time for A & B .

c. At what time(s), if any, does A pass B?

d. At what time(s), if any, does B pass A?

1 s, 3 s

3 s

1 s

v

t

v

t

A B

2.3 motion with constant acceleration

Given:

derive: 1. vx = vx0 + axt

(assume t0 = 0)

2. x = x0 + vx0 + ½ axt2

3. vx2 – vx0

2 = 2ax(x – x0)

Homework – extra credit

t

xxvxav

0

20xx

xav

vvv

t

vva xx

xav0

Motion with constant acceleration vx-t graph

A horizontal line indicate the slope = 0, a = 0

Since ax = ∆v / ∆t; ∆v = ax∙∆t which is represented by the area.

a-t graph

The area indicate the change in velocity during ∆t

Kinematics equations for constant acceleration

Example 2.4A 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 2.5A 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 2.4Four possible vx-t graphs are shown for the two vehicles in example 2.5. which graph is correct?

example

10

0

20

-10

-20

5 10 14

1. At what time after t = 0 does the object again pass through its initial position?

2. During which interval does the particle have the same average acceleration as 12 s < t < 14 s?

a. 9s < t < 11s

b. 2s < t < 5s

c. 0s < t < 3s

d. 3s < t < 7s

e. 5s < t < 11s

9 s

v (m/s)

Example 2.22

• 1 mi/hr = 0.447 m/s 173 mi/h = 77.3 m/s• 1 ft = 0.3048 m 307 ft = 93.6 m

a. ax = 32.0 m/s2

b. t = 2.42 s

The catapult of the aircraft carrier USS Abraham Lincoln accelerates an F/A-18 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.

Examplev

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?

t

x

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

2.5 Free Falling Bodies

Example 2.6A 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 2.7You 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. finda.The position and velocity of the ball 1.00 s and 4.00 s after leaving your handb.The velocity when the ball is 5.00 m above the railingc.The maximum height reached and the time at which it is reachedd.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 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?

4h

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)

• 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

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? 6 m/s2

• 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 x-velocity. And we can use ax = dvx/dt to find the x-acceleration 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).

2.6 velocity and position by integrationFinding v(t) and x(t) when given a(t)

The “AREA”In v-t graph, the area under the line represent displacement.

How ever, if acceleration is not constant, how can we determine x(t)?

t1 t2

v (m/s)

t(s)

x = area

t1 t2

v (m/s)

t(s)

x = area

t1

v (m/s)

t(s)t1

v (m/s)

t1

v (m/s)

v(t)

t2

v (m/s)

Zoom in

dt

dxv

t

xv t

0limWe have learned that the rate of change of displacement is defined as the VELOCITY of an object. Consider the graph below

dttvdx )(

t1 t2

v (m/s)

t(s)tt1 t2

v (m/s)

t(s)tt1 t2

v (m/s)

t(s)tt1

v(t)

t2

v (m/s)

t(s)t

dt

∆Area = v(t)∙dt

TOTAL DISPLACEMENT = ∑∆Area = ∑v(t)∙dt

The “Integral” – the area• The temptation is to use the conventional summation sign “" .

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?

dttdtvtx )2()(

2)( ttx

44927

)2()(

22

272

7

2

7

2tdttdtvtx t

t

t

t

t

t

These are your LIMITS!

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.

dtavdtvx

dt

dva

dt

dxv

ExampleHere is a simple example of which you

may be familiar with:Assume we know the circumference

of a circle is 2r, where r is the radius. How can we derive an

expression for the area of a circle whose radius is R?

We begin by taking a differential 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

∫xn dx = + Cxn+1

n+1 ∫xndx = -bn+1

n+1a

b an+1

n+1

ddx (xn+1) = (n+1) xn

General expression, no limits

Evaluation with given limits

Special case: n = 0 ∫ dx = x + C ∫a

b

dx = b - aor

Addition/subtraction ruleconstant 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 x-acceleration is a function of time:

a. find her x-velocity and position as functions of time

b. When is her x-velocity greatest?

c. What is the maximum x-velocity?

d. Where is the car when it reaches the maximum x-velocity?

tsmsmax )/10.0(/0.2 32

Position, velocity and acceleration of the car

Example 2.10

To find vx and x as functions of time in the case in which the acceleration is constant.

t

xo

t

xxx

dtvxx

dtavv

0

0

0

Use Eqs.

ExampleA particle moving in one dimension has a position function

defined as: x(t) = 6t4-2ta) 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?

0)26( 4

dt

ttd

dt

dxva)

b)

t = 0.437 s

12)224( 3

dt

td

dt

dva t = 0.408 s v = -0.37 m/s

Example 2.50The 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 = vox + (2 m/s3)½ t2

vox = 0

vx = (1 m/s3)t2

vx = (1 m/s3)t2

x = xo + (1 m/s3)(1/3)t3

x(3 s) – x (0 s) = (1 m/s3)(1/3)(3 s)3

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