Homework – due Friday, 9/23Reading assignment: 2.1-2.6

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 v2v2

A B C D E

v2

-v1

v2 -v1

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

• 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 the time interval ∆t . vav-x = =

∆x

∆t

x2 – x1

t2 – t1

∆x Displacement can be zero, positive or negative.

-positive if you move to the positive direction-negative if you move to the negative direction.

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

• 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.

+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

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)

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

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,

2. You average velocity from pillar to post.

4.4 m/s

-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?

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

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.

c. tRick = (D ∕ 2)/vr + (D/2)/ vw

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)

solution

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.

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.

• 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.

P1P2

• In the language of calculus, the limit of as ∆t approaches zero is called the derivative of x with the respect to t and is written as dx

dt

∆x∆t

vx = lim = dx dt

∆x ∆t∆t 0

When ∆x is positive, vx is positive

When ∆x is negative, vx is negative

Example 2.1• 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.

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

Common derivative equations

ddx

C = 0

ddx

Cu(x) =

Cdudx

=

nxnddx

xn-1

=

sinxddx

cosx =

cosxddx

-sinx

=

exddx

ex =

lnxddx

1x

ddx

(u+v) =

dudx

dvdx+

y = g(u); u = f(x) dudx=

dydx •

dydu

Derivatives Practice• Find the derivatives (dx/dt) of the

following function

1. x = t3

2. x = 1/t

3. x = 6t3 + 2/t

4. 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

• 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

velocity on P-T graphSlope of the line P1P2 represents the average velocity v between t1 and t2.

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

t

x

0

A

B

C

D E

The slope of the tangent at any point equals the velocity at that point.

B

1. At which point does the particle has greatest positive velocity?

2. At which point does the particle has zero velocity? C

3. At which point does the particle has smallest nonzero negative velocity?

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

Q

R

S

P

R

Q, S

R, P, Q = 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

Constant (uniform) motion

2.80

x

t

t

v

2.80

Average velocity = instantaneous velocity

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.

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.

aav-x = =v2 – v1

t2 – t1

∆v

∆t

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.

aav-x = =v2 – v1

t2 – t1

∆v

∆t

aav-x = lim∆v

∆tt 0

aav-x =dv

dtSince v =

dx

dt

d2x

dt2

dv

dt=

d2x

dt2=aav-x

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 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. 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 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

• 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

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 graphOn 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

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

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.

Derive equations for motion with constant acceleration

Given: vxav =x – x0

tvxav =

vx + vxo

2

axav =vx – vx0

t

derive: 1. vx = vxo + axt

(assume t0 = 0)

2. x = xo + vxo + ½ axt2

3. vx2 – vxo

2 = 2ax(x – x0)

Objectives: Motion with constant velocity

ax

t

vx

t

ax

t

x

t

Graphs of motion

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

a-t graph

Slope: indicate acceleration

Area indicate the change in displacement from time = 0 to t

v-t graph

Kinematics equations for constant acceleration

Example 2.4

a.

b. X = 55 m

a. t = 10 s. b. v = 30 m/s. c. d = 150 m

Example: 2.5

Test your understanding 2.4

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

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

An example of constant acceleration – acceleration due to gravity

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) – ½ (9.80 m/s/s)t2

(9.80 m/s/s)t2 – (15 m/s)(t) – (5.00 m) = 0

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

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

timeT 2T

9/30 Do nowv

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

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

• 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

10/4 do now

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? 6 m/s2