1-d Motion: Position & Displacement

R. Field 1/17/2013 University of Florida

PHY 2053 Page 1

1-d Motion: Position & Displacement

We locate objects by specifying their position along an axis (in this case x-axis). The positive direction of an axis is in the direction of increasing numbers. The opposite is the negative direction.

• The x-axis:

• Displacement: The change from position x1 to position x2 is called the displacement, x.

x = x2 –x1

• Graphical Technique:

The displacement has both a magnitude, |x|, and a direction (positive or negative).

A convenient way to describe the motion of a particle is to plot the position x as a function of time t (i.e. x(t)).

time t

x(t)


PHY 2053 Page 2

1-d Motion: Average Velocity

• Average Velocity 12

12

12

12 )()(

tt

xx

tt

txtx

t

xvv ave

The average velocity is defined to be the displacement, x, that occurred during a particular interval of time, t (i.e. vave = x/t).

• Average Speed The average speed is defined to be the magnitude of total distance covered during a particular interval of time, t (i.e. save = (total distance)/t).


PHY 2053 Page 3

1-d Motion: Instantaneous Velocity

t t+tt

x(t+t)

x(t)

shrink t

t

x


PHY 2053 Page 4


t t+tt

x(t+t)

x(t)

shrink t

t

x


PHY 2053 Page 5


t t+t

t

x(t+t)

x(t)

shrink t

t

x


PHY 2053 Page 6


t t+t

t

x(t+t)x(t)

shrink t

t

x


PHY 2053 Page 7


t

x(t) Instantaneous velocity v(t) is slope of x-t tangent line at t

tangent line at t

t

x

dt

dx

t

xv

t

0

lim The velocity is the derivative of x(t) with respect to t.


PHY 2053 Page 8

1-d Motion: Acceleration

Hello

dt

dv

t

va

t

0

lim

• Average Acceleration The average acceleration is defined to be the change in velocity, v, that occurred during a particular interval of time, t (i.e. aave = v/t).

• Instantaneous Acceleration The acceleration is the derivative of v(t) with respect to t.

• Acceleration When a particles velocity changes, the particle is said to undergo acceleration (i.e. accelerate).

v1

v2

vv(t)

v “rise”

av

12

12

12

12 )()(

tt

vv

tt

tvtv

t

vaa ave

v

v(t)Instantaneous acceleration a(t) is

slope of v-t tangent line at t


PHY 2053 Page 9

Equations of Motion: a = constant

221

00 attvxx

ata )(

)(2 020

2 xxavv

atvv 0

• Special case! (constant acceleration)

• v is a linear function of t

• x is a quadratic function of t

v at t = 0

x at t =0

Velocity as a function of time t: v(t)

0

5

10

15

20

0 1 2 3 4 5 6 7 8 9 10time t (in s)

Vel

oci

ty v

(in

m/s

)

a = 2 m/s2

v0 = 0

Position as a function of time t: x(t)

0

20

40

60

80

100

120

0 1 2 3 4 5 6 7 8 9 10

time t (in s)

Po

siti

on

x (

in m

)

a = 2 m/s2

v0 = 0x0 = 0

Acceleration as a function of time t: a(t)

0

1

2

3

4

5

0 1 2 3 4 5 6 7 8 9 10time t (in s)

Acc

eler

atio

n a

(in

m/s

2)

a = 2 m/s2

• Note also that


PHY 2053 Page 10

Acceleration Due to Gravity

2/8.9 smgay

gtvtv yy 0)(

221

00)( gttvyty y

• Experimental Result

Earth

y-axis

x-axis

RE

h

Near the surface of the Earth all objects fall toward the center of the Earth with the same constant acceleration, g ≈ 9.8 m/s2, (in a vacuum) independent of mass, size, shape, etc.

• Equations of Motion

)(2 02

02 yygvv yy

The acceleration due to gravity is almost constant

and equal to 9.8 m/s2 provided h << RE!


PHY 2053 Page 11

Equations of Motion: Example Problem

smvv yy /490

ssm

sm

g

vt y 5

/8.9

/492

0

• Example Problem A ball is tossed up along the y-axis (in a vacuum on the Earth’s surface) with an initial speed of 49 m/s.

msm

sm

g

vh y 5.122

)/8.9(2

)/49(

2 2

220

y-axis

vy0 = 49 m/s

EarthHow long does the ball take to reach its maximum height?

What is the ball’s maximum height?

Velocity as a function of time t: v(t)

-60

-40

-20

0

20

40

60

0 1 2 3 4 5 6 7 8 9 10time t (in s)

Vel

oci

ty v

(in

m/s

)

a = -9.8 m/s2

v0 = 49 m/s

Position as a function of time t: y(t)

0

20

40

60

80

100

120

140

0 1 2 3 4 5 6 7 8 9 10

time t (in s)P

osi

tio

n y

(in

m)

a = -9.8 m/s2

v0 = 49 m/sy0 = 0

How long does it take for the ball to get back to its starting point?

sg

vt y 10

2 0

What is the velocity of the ball when it gets back to its starting point?


PHY 2053 Page 12

2-d Motion: Constant Acceleration

yaxaa yx ˆˆ

tavtv

0)(

yvxvv yx ˆˆ 000

• Kinematic Equations of Motion (Vector Form)

The velocity vector and position vector are a function of the time t.

221

00)( tatvrtr

Acceleration Vector (constant)

Velocity Vector (function of t)

Position Vector (function of t)

yyxxr ˆˆ 000

Velocity Vector at time t = 0.

Position Vector at time t = 0.

The components of the acceleration vector, ax and ay, are constants.

The components of the velocity vector at t = 0, vx0 and vy0, are constants.

The components of the position vector at t = 0, x0 and y0, are constants.

Warning! These equations are only valid if the acceleration is constant.


PHY 2053 Page 13

2-d Motion: Constant Acceleration

xatavtv xxx 0)(

221

00)( tatvxtx xx

• Kinematic Equations of Motion (Component Form)

constant

The components of the acceleration vector, ax and ay, are constants.

The components of the velocity vector at t = 0, vx0 and vy0, are constants.

The components of the position vector at t = 0, x0 and y0, are constants.

Warning! These equations are only valid if the acceleration is constant.

yatavtv yyy 0)(

221

00)( tatvyty yy

constant

• Ancillary Equations

))((2)( 022 xtxavtv xxox

))((2)( 022 ytyavtv yyoy

Valid at any time t


PHY 2053 Page 14

Example: Projectile Motion

cos)( 0vtvx tvtx )cos()( 0

• Near the Surface of the Earth (h = 0) In this case, ax= 0 and ay= -g, vx0 = v0cos, vy0 = v0sin, x0 = 0, y0 = 0.

gtvtvy sin)( 0

221

0 )sin()( gttvty

x-axis

y-axis

v0

• Maximum Height H The time, tmax, that the projective reaches its maximum height occurs when vy(tmax) = 0. Hence,

g

vt

sin0max

g

vtyH

2

)sin()(

20

max

• Range R (maximum horizontal distance traveled) The time, tf, that it takes the projective reach the ground occurs when y(tf) = 0. Hence,

221

0 )sin()(0 fff gttvty g

vt f

sin2 0

g

v

g

vtvtxR ff

2sincossin2)cos()(

20

20

0


PHY 2053 Page 15

Example: Projectile Motion

cos)( 0vtvx tvtx )cos()( 0

• Near the Surface of the Earth (h = 0) In this case, ax= 0 and ay= -g, vx0 = v0cos, vy0 = v0sin, x0 = 0, y0 = 0.

gtvtvy sin)( 0

221

0 )sin()( gttvty

x-axis

y-axis

v0

• Maximum Height H The time, tmax, that the projective reaches its maximum height occurs when vy(tmax) = 0. Hence,

g

vt

sin0max

g

vtyH

2

)sin()(

20

max

• Range R (maximum horizontal distance traveled) The time, tf, that it takes the projective reachthe ground occurs when y(tf) = 0. Hence,

221

0 )sin()(0 fff gttvty g

vt f

sin2 0

g

v

g

vtvtxR ff

2sincossin2)cos()(

20

20

0

For a fixed v0 the largest R occurs when = 45o!


PHY 2053 Page 16

Exam 1 Fall 2012: Problem 11

d

h

ground

building

v0 • Near the surface of the Earth a projectile is fired from the top of a building at a height h above the ground at an angle relative to the horizontal and at a distance d from the edge of the building as shown in the figure. If = 20o and d = 20 m, what is the minimum initial speed, v0, of the projectile such that it will make it off the building and reach the ground? Ignoring air resistance.

Answer: 17.5 m/s % Right: 35%

g

vRd

2sin20

smsmsmmdg

v /5.17/46.17)202sin(

)/8.9)(20(

2sin

2

0


PHY 2053 Page 17

Exam 1 Spring 2012: Problem 12

• A beanbag is thrown horizontally from a dorm room window a height h above the ground. It hits the ground a horizontal distance d = h/2 from the dorm directly below the window from which it was thrown. Ignoring air resistance, find the direction of the beanbag's velocity just before impact.Answer: 76.0° below the horizontal% Right: 22%

h

d

x-axis

y-axis

tvtx

vtvx

0

0

)(

)(

Let th be the time the beanbag hits the ground.

221)(

)(

gthty

gttvy

76

42

)(

)(tan

2

0

d

h

d

gt

v

gt

tv

tvhh

hx

hy

hh

hh

tvdtx

gthty

0

221

)(

0)(

h

h

t

dv

g

ht

0

2 2


PHY 2053 Page 18

Example Problem: Projectile Motion

0)( tvx0)( tx

• A suspension bridge is 60.0 m above the level base of a gorge. A stone is thrown or dropped off the bridge. Ignore air resistance. At the location of the bridge g has been measured to be 9.83 m/s2. If you drop the stone how long does it take for it to fall to the base of the gorge?

In this case, ax= 0 and ay= -g, vx0 = 0, vy0 = 0, x0 = 0, y0 =h. Hence,

gttvy )(2

21)( gthty

The time, tf, that the it takes the stone to reach the ground occurs when y(tf) = 0. Hence,

x-axis

y-axis

h

2210)( ff gthty s

sm

m

g

ht f 494.3

)/83.9(

)60(222

• If you throw the stone straight down with a speed of 20.0 m/s, how long before it hits the ground?

In this case, ax= 0 and ay= -g, vx0 = 0, vy0 = -v0, x0 = 0, y0 =h. Hence,

221

0)( gttvhty

ssm

msmsmsm

g

ghvvt of 01.2

/83.9

)60)(/83.9(2)/20(/202)(2

2220

Have to use the Quadratic Formula!

221

00)( fff gttvhty

1-d Motion: Position & Displacement

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