Chapter 2: Motion in One Dimension

(1)

PHU 205

Mechanics for Life Sciences

Chapter 2:

Motion in One Dimension

Prof. Liliana Braescu &

(2)

Required text

College Physics

Raymond A. Serway and Chris Vuille 9th_{Edition, 2012}

BROOKS/COLE CENGAGE Learning ISBN 10: 1-111-42745-3

ISBN 13: 978-1-111-42745-0

(3)

Course Outline

Chapter 2

Motion in One Dimension

2.1 Displacement 2.2 Velocity

2.3 Acceleration

2.4 Motion Diagrams

2.5 One Dimensional Motion with Constant Acceleration

(4)

Terminology

u Mechanics—study of motion and change of position of an object

u Kinematics—description of motion in terms of space-time without regard to the causes of motion (x, v, a, t)

u Dynamics—study of the physical causes of motion and interplay between motion, forces, and properties of moving objects (x, v, a, t, & F)

(5)

Quantities in Motion

u Any motion involves three concepts ✔ Displacement

✔ Velocity

✔ Acceleration

u These concepts can be used to study objects in motion

(6)

Chapter 2:

2.1 Displacement

2.2 Velocity

2.3 Acceleration

2.4 Motion Diagrams

(7)

6

2.1 Displacement

u The displacement of an object is the change

in its position.

u It is defined by coordinates on a straight line:

u Displacement can be positive or negative.

u It can also be zero if there is no change in position (i.e. the object begins and ends in the same place).

i f x x x Δ = −

(8)

Displacement - Examples

uFrom A to B ✔ x_i = 30 m ✔ x_f = 52 m ✔ Δx = 22 m T h e d i s p l a c e m e n t i s positive => motion was in the positive x direction

uFrom C to F

✔ x_i = 38 m ✔ x_f = -53 m ✔Δx = -91 m

T h e d i s p l a c e m e n t i s negative => motion was in

i f

x x x

(9)

Displacement – Graphical representation

uFrom A to B ✔ x_i = 30 m ✔ x_f = 52 m uFrom C to F ✔ x_i = 38 m i f x x x Δ = −

(10)

Remark:

Displacement is not Distance

u The displacement of an object is not the same

as the distance it travels

Example: Throw a ball straight up and then catch it at the same point you released it.

✔ The distance is twice the height (distance

is always positive!)

✔ The displacement is zero (displacement

can be positive, negative or zero!)

(11)

Vectors and Scalar Quantities

u Vector quantities need both magnitude (size)

and direction to completely describe them

u Generally denoted by boldfaced type and an arrow over the letter

+ or – sign is sufficient for this chapter

u Scalar quantities are completely described by magnitude only

(12)

Chapter 2:

2.1 Displacement

2.2 Velocity

2.3 Acceleration

2.4 Motion Diagrams

(13)

2.2 Velocity

u Average velocity is a measure of displacement divided by the time interval over which the displacement occurs:

u On a graph of position versus time, velocity equals the slope of the line.

v

!"

= Δ

x

Δt

=

x

_f

− x

_i

t

_f

− t

_i

(14)

Velocity

u Direction will be the same as the direction of the displacement, + or - is sufficient

u Units of velocity are m/s (SI)

✔ Other units may be given in a problem, but generally will need to be converted to these ✔ In other systems:

- US Customary: ft/s - cgs: cm/s

(15)

Speed

u The average speed of an object is defined as the total distance traveled divided by the total time elapsed

u Speed is a scalar quantity

=

total distance

Average speed

total time

d

v

t

(16)

Speed vs. Velocity

u Average speed is a measurement of the total distance traveled over the time interval.

(17)

Speed vs. Velocity

u Cars on both paths have the same average velocity since they had the same displacement in the same time interval

u The car on the blue path will have a greater average speed since the distance it traveled is larger

(18)

Graphical Interpretation of Velocity

u Velocity can be determined from a position-time graph

u Average velocity equals the slope of the line joining the initial and final positions

u An object moving with a constant velocity will have a graph that is a straight line

(19)

Average Velocity, Constant

u The straight line indicates constant velocity

u The slope of the line is the value of the average velocity

(20)

Remarks:

u The general equation for the slope of any line is

u Slope carries units

u The meaning of a specific slope will depend on the physical data being graphed

change in vertical axis

slope

change in horizontal axis

(21)

Average Velocity,

Non-Constant

u The motion is non-constant velocity

u The average velocity is the slope of the straight line joining the initial and final points

(22)

Instantaneous Velocity

u Instantaneous velocity is the velocity at a given time:

u Instantaneous speed is the absolute value of the instantaneous velocity.

u Consider a traveling car…

✔ Its average speed is found between two different points of the trip.

✔ Its instantaneous speed is found at one point in time (almost equal to reading the speedometer) 0 lim t x v t Δ → Δ = Δ

(23)

Graphical Instantaneous Velocity

u Average velocities are the blue lines

u The green line (tangent) is the instantaneous velocity

(24)

Problems

(for more examples see the Book - Chapter 2)

u  Problem 2.1

An athlete swims the length of a 50.0-meter pool in 20.0 seconds and makes the return trip to the starting position in 22.0 seconds.

Question: Determine her average velocities in (a) the first half of the swim,

(b) the second half of the swim, and (c) the round trip.

(25)

Problems

Solution

(a) In the first half of the trip,

velocity = (x₂ – x₁)/20.0 s

= +50.0 m/20.0 s = +2.50 m/s

(b) On the return leg,

velocity = (x₃ – x₂)/22.0 s

= (0 - 50.0 m)/22.0 s = -2.27 m/s

(c) For the entire trip (time =20+22 s), velocity = (x₃ – x₁)/42.0 s = 0/42.0 s = 0 m/s v !" = Δx Δt = x_f − x_i t_f − t_i

(26)

Problems

A tennis player moves in a straight-line path as shown in the figure.

Question: Find her average velocities in the time intervals

(a) 0 to 1.0 second, (b) 0 to 4.0 seconds,

(c) 1.0 to 5.0 seconds, and (d) 0 to 5.0 seconds.

(27)

Problems

u  Problem 2.2 Solution (a) v = (x₁ – x₀)/∆t = (4.0 m - 0)/1.0 s = +4.0 m/s (b) v = (x₄ – x₀)/∆t = (-2.0 m - 0)/4.0 s = -0.5 m/s (c) v = (x₅ – x₁)/∆t = (0 - 4.0 m)/4.0 s = -1.0 m/s (d) v = (x₅ – x₀)/∆t = (0 – 0 m)/5.0 s = 0 m/s v !" = Δx Δt = x_f − x_i t_f − t_i (a) 0 to 1.0 second, (b) 0 to 4.0 seconds, (c) 1.0 to 5.0 seconds (d) 0 to 5.0 seconds.

(28)

Problems

Find the instantaneous velocities of the tennis player in problem 2.2 at (a)  0.50 seconds, (b)  2.0 seconds, (c) 3.0 seconds, and (d) 4.5 seconds. Hint:

Instantaneous velocity (tangent)

x

(29)

u  Problem 2.3 Solution (a) v(t = 0.50 s) = [x(t = 1 s) - x (t = 0)] /(1.00 s) = +4.00m/1.00s = +4.00 m/s (b) v(t = 2 s) = [x(t = 2.5 s) - x(t = 1 s)] /(2.50 s - 1.00 s) = (-2.0 m - 4.0 m)/1.5 s = -6.0 m/1.5 s = -4.0 m/s (c) v(t=3 s) = [x(t = 4 s) - x(t = 2.5 s)]/(4.0s - 2.5 s) = (-2.0m -(-2.0m))/1.5 s = 0 m/s (d) v(t = 4.5 s) = [x(t = 5 s) - x(t = 4s)]/(5 s - 4 s) = (0 - (-2.0 m))/1.0 s = +2.0 m/s 0 lim t x v t Δ → Δ = Δ

(30)

Problems

If the average speed of an orbiting space shuttle is 19,800 mph, determine the time required for it to circle the Earth.

Make sure you consider the fact that the shuttle is orbiting about 200 miles above the Earth’s surface, and assume that the Earth’s radius is 3963 miles.

(31)

Problems

Solution

The distance traveled by the space shuttle in one orbit equals

Δx = 2π(Earth's radius + 200 miles) = 2π(3963 + 200)

= 26,156.9 miles.

So the required time is given by

Δt = 26,156.9 miles/19800 mph

(32)

Chapter 2:

2.1 Displacement

2.2 Velocity

2.3 Acceleration

2.4 Motion Diagrams

(33)

2.3 Acceleration

u When velocity changes with time, the object undergoes acceleration.

u If velocity does not change with time, acceleration is equal to zero and the motion is said to be uniform (constant).

u Average acceleration is a measure of change in velocity divided by the time interval over which the change occurs f i i f v v v a t t t − Δ = = Δ −

(34)

Average Acceleration

u When the object s velocity and acceleration are in the same direction (either positive or negative), then the speed of the object increases with time

u When the object s velocity and acceleration are in the opposite directions, the speed of the object decreases with time

u Vector quantity

(35)

Relationship: Acceleration and Velocity

! Uniform velocity (shown by red arrows maintaining the same size)

(36)

Relationship: Acceleration and Velocity

u Velocity and acceleration are in the same direction

u Acceleration is uniform (violet arrows maintain the same length)

u Velocity is increasing (red arrows are getting longer)

(37)

Relationship: Acceleration and Velocity

u Acceleration and velocity are in opposite directions

u Acceleration is uniform (violet arrows maintain the same length)

u Velocity is decreasing (red arrows are getting shorter)

(38)

Instantaneous vs. Average Acceleration

u Instantaneous acceleration is the acceleration at a given time (limit from average acceleration):

u Instantaneous acceleration is the slope of the tangent to the curve of the velocity-time graph

u Average acceleration is the slope of the line connecting the initial and final velocities on a velocity vs. time graph

0

lim

t

v

a

t

Δ →

Δ

=

Δ

(39)

Instantaneous

vs.

Average Acceleration

Slope of the tangent to the Curve vs.

Slope of the Line Connecting Initial and Final Points

(40)

Problems

A certain car is capable of accelerating at a rate of +0.60 m/s2_.

Question: How long does it take for this car to go

(41)

Problems

Solution

u Begin by converting initial and final velocities to SI units:

v_i = 55 mph = 24.58 m/s

v_f = 60 mph = 26.82 m/s. u Thus, Δv = 2.24 m/s, and

Δt = Δv/a

= (2.24 m/s)/(0.6 m/s2)

(42)

Chapter 2:

2.1 Displacement

2.2 Velocity

2.3 Acceleration

2.4 Motion Diagrams

(43)

2.4 Motion Diagrams

Summary

(44)

2.4 Motion Diagrams

Summary

u If a = 0, velocity is constant (uniform motion).

u An object is speeding up whenever its velocity and acceleration have the same sign.

u If the velocity and acceleration have different signs, the object is slowing down.

u Four scenarios are possible:

v > 0 and a > 0

→ object is moving to the right, speeding up

v > 0 and a < 0

→ object is moving to the right, slowing down v < 0 and a > 0

→ object is moving to the left, slowing down v < 0 and a < 0

(45)

Chapter 2:

2.1 Displacement

2.2 Velocity

2.3 Acceleration

2.4 Motion Diagrams

2.5 One Dimensional Motion with Constant Acceleration

(46)

2.5 One Dimensional Motion with

Constant Acceleration

u When a is constant (not equal to zero), four relations can be derived between position, velocity, acceleration, and time.

u Originally derived by Galileo, these are the equations of kinematics.

u Any kinematical problem can be solved using one or more of these.

(47)

2.5 One Dimensional Motion with

Constant Acceleration

(48)

2.5 One Dimensional Motion with

Constant Acceleration

(49)

2.5 One Dimensional Motion with

Constant Acceleration

(50)

First Equation

u

Take t

_i

=0, t

_f

=t, v

_i

=

v

₀

, v

_f

=v

a = (v -

v

₀

)/t

we have

v = v

₀

+ at

(1)

(51)

Derivation of Kinematics Equations

Second Equation

u Since v =

v

₀

+ at ,

u

Total area under the v—t curve

x= v

₀

t+(v-v

₀

)t /2

and hence

(52)

Derivation of Kinematics Equations

Third Equation

u Inserting v = v

₀

+ at into Eq(2), we

can write the displacement as

x = (v+v

₀

)t /2 = [(v

₀

+ at) + v

₀

] t /2

or

(53)

Derivation of Kinematics Equations

Fourth Equation

u Now, since t = (v-v

₀

)/a

u Eq. (2) leads to

x = (v+v

₀

)t /2 = (v+v

₀

) (v-v

₀

)/2a

=( v

2

_-v

02

)/2a

which in turn leads to

v

2

_-v

(54)

Conclusion: Kinematics Equations are

uIf acceleration: a = constant, then

v = v₀ + at (1) x = (v+v₀)t /2 (2) x = v₀ t + at2_/2 ₍₃₎ v2_-v 02= 2ax (4)

(55)

Problems

A Cessna aircraft has a lift-off speed of 120 km/h.

Questions:

(a) What minimum constant acceleration does this require if the aircraft is to be airborne after a takeoff run of 240 meters?

(b) How long does it take the aircraft to become airborne?

(56)

Problems

u  Problem 2.6 Solution (a) With 120 km/h = 33.33 m/s, v2_{= v} 02 + 2ax (33.33 m/s)2_{= 0 + 2a(240 m).} => a = 2.32 m/s2_. (b) Using v = v₀ + at, 33.33 m/s = 0 + (2.32 m/s2_)t, or t = 14.4 s.

(57)

Problems

A jet plane lands with a velocity of +100 m/s and can accelerate at a maximum rate of -5.0 m/s2_as

it comes to rest.

Questions:

(a) From the instant it touches the runway, what is the minimum time needed before it can come to rest?

(b) Can this plane land on a small island airport where the runway is 0.80 km long?

(58)

Problems

u  Problem 2.7 Solution (a) t = (v – v₀)/a = (0 – 100 m/s)/-5 m/s2 = 20 s (b) x = vt = 50 m/s * 20 s = 1000 m

Note that the minimum distance to stop is 1 km, which exceeds the length of the runway.

(59)

Chapter 2:

2.1 Displacement

2.2 Velocity

2.3 Acceleration

2.4 Motion Diagrams

(60)

2.6 Freely Falling Objects

u When an object is moving under the influence of gravity alone—regardless of its initial conditions—it is experiencing free fall motion.

✔ Ini(al velocity is zero

✔ Once in free fall, all objects experience a downward

acceleration equal to the acceleration due to gravity: -g = -‐9.80 m/s2

✔The equations of kinematics can be rewritten to describe free fall motion by replacing –g for a and y for x.

v_o= 0

(61)

Free Fall Equations

u When air resistance is neglected, we have v = v₀- gt (1) y = (v+v₀ )t /2 (2) y = v₀t – gt2_/2 ₍₃₎ v2_-v 02 = - 2gy (4) On earth: g = 9.8 m/s2

Note: v₀can be (+) or (-) depending on the motion being upward or downward.

(62)

Remarks:

u When an object is free-falling, its speed increases rapidly with every passing second.

u Acceleration due to gravity varies on different stellar bodies:

g(moon) = 1.6 m/s2

g(sun) = 270 m/s2

g(neutron star) = 1.3 x 1013_m/s2

g(black hole) = ∞

u  Neglecting air resistance, all objects fall

at the same rate.

(63)

Problems

A ball thrown vertically upward is caught by the thrower after 2.00 seconds

Questions: Find

(a) the initial velocity of the ball and (b) the maximum height it reaches.

(64)

63

Problems

Solution

(a) Using a = -g and placing the origin at the thrower yields y = v₀t – gt2_/2.

✔ When y = 0, we get [v₀ - gt/2]t = 0, which has two solutions, t = 0, and t = 2v₀/g.

✔The t = 0 solution corresponds to when the ball is thrown, while the t = 2v₀/g solution corresponds to when the ball is caught.

✔Therefore, the initial velocity must be

v₀= gt/2 = (9.80)(2.00)/2 = 9.80 m/s. (b) Using v2_{= v}

02 - 2gy, we have, at the maximum

height, v = 0, so that

(65)

Problems

A parachutist with a camera, both descending at a speed of 10 m/s, releases that camera at an altitude of 50 meters.

Questions:

(a) How long does it take the camera to reach the ground?

(b) What is the velocity of the camera just before it hits the ground?

(66)

Problems

Solution 1

(a) Choose the origin of the coordinate system at the location of the parachutist.

✔ We use v2_{= v}

02 + 2ay, which gives

v2_{= (-10 m/s)}2_{+ 2(-9.80 m/s}2_{)(-50 m),}

from which v = -32.86 m/s.

✔The time to reach the ground is found from

v = v₀ + at.

This yields -32.86=-10-9.8×t and hence t= 2.33 s.

(b) The velocity was found to be -32.86 m/s in the previous part (a).

(67)

Problems

Solution 2 (used during the lecture)

(a) Choose the origin of the coordinate system at the location of the parachutist.

✔ We use y = v₀t – gt2_{/2, which gives}

-50 m=(-10 m/s)t-(9.8 m/s2_{) (t}2_/2),

from which 9.8t2_{+20t-100=0 and hence t}

1=2.33 s,

t₂=-4.37 s. The value t= 2.33 s has physical meaning. (b) The velocity is found from

v = v₀ + at.

This yields v=(-10 m/s)-(9.8 m/s2_{)×(2.33 s) and hence}

(68)

Chapter 2 Ends

Quiz 2 is coming …..

before Chapter 3 will start

For review,

Solve the following problems: 24, 25 page 51;

32, 33 page 52; and 51, 54 page 53.

Textbook “College Physics”

Raymond A. Serway and Chris Vuille 9th_{Edition, 2012}