Chapter 2: Motion in One Dimension

Introduction The study of motion and of physical concept such as force and mass is called dynamics.

The part of dynamics that describes motion without regard to its causes is called kinematics.

In this chapter, we will learn about kinematics in one dimension: motion along a straight line.

Motion involves the displacement of an object from one place to another. For that we need a convenient coordinate system and a specific origin : A frame of reference

Displacement

0

x (m)

displacement x:

if xxx

finalposition

initialposition

(origin)

Displacement:

Displacement

0

if xxx

(origin)

Point A to point B: m 22m 52 m, 30 xxx fi

Point C to point F: m 91m 53 m, 38 xxx fi

• Displacement has both a magnitude and a direction: It’s a vector.

• A quantity that has both a magnitude and a direction is called a vector.

• A quantity that has only a magnitude and no direction is called a scalar.

Velocity vs. speed

Velocity

Speed is a scalar quantity, while velocity is a vector quantity.

• Average speed of an object over a given time interval:

• Average velocity of an object over a given time interval:

• Average velocity vs. average speed

timetotal

distance totalspeedAverage SI unit: meter per second m/s

if

if

tt

xx

t

xv

SI unit: meter per second m/s

average speed = vaverage speed = v

__

Velocity

Point A to point B:

s 10s 10 s, 0

m 22m 52 m, 30

ttt

xxx

fi

fi

Point C to point F:

s 30 s 50 s, 20

m 91m 53 m, 38

ttt

xxx

fi

fi

Average velocity examples

m/s 2.2s m/10 22 v

m/s 0.3s m/30 91 v

Velocity

Point A to point B:

s 10s 10 s, 0

m 22m 52 m, 30

ttt

xxx

fi

fi

Graphical interpretation of velocity

m/s 2.2s m/10 22 v slope

Average velocity

Instantaneous velocity

dt

dx

t

xv

t

0

lim

derivative of x with respect to tat point A or slope of line tangentto point A

Velocity

Graphical interpretation of velocity

Acceleration

Average acceleration

if

if

ttt

vvv

change of velocity

change of time

if

if

tt

vv

t

va

average acceleration SI unit: m/s2

Acceleration

Average acceleration (cont’d)

s )02(

m/s )1020(

if

if

ttt

vvvchange of velocity

change of time

2m/s 5s 2

m/s 10

t

vaaverage acceleration

=0.0 s=2.0 s=10 m/s

=20 m/s

Acceleration

Instantaneous acceleration

dt

dvt

va

t

0

liminstantaneous acceleration

The Instantaneous accelerationof an object at a given time equalsthe slope of the tangent to thevelocity vs. time graph at that time.

Examples• Example 2.2: Slowly moving train

a) Average velocity from O to C:

m/s 833.0s 12.0

m 0.10

t

xv

b) Average velocity from O to A:

m/s 00.1s 4.00

m 00.4

t

xv

d) Instantaneous velocity at t=2.00 s:

m/s 00.1v

d) Instantaneous velocity at t=9.00 s:

m/s 75.0s 0.3s 0.9

m 0m 5.4

t

xv

3.0 s

9.0 s

4.5 m

Examples of Velocity and Acceleration

Examples of Velocity and Acceleration

Examples• Example 2.3: Catching a fly ball

a) Instantaneous acceleration at A:

2m/s 0.2s 0.0s 0.2

m/s 0.0 m/s 0.4

t

va

b) Instantaneous acceleration at B:

2m/s 0.0s 0.2s 0.3

m/s 0.4 m/s 0.4

t

va

c) Instantaneous acceleration at C:

2m/s 0.2s 0.3s 0.4

m/s 0.4 m/s 0.2

t

va

One-dimensional Motion with Constant Acceleration

Instantaneous velocity• When an object moves with constant acceleration, the instantaneous acceleration at any point in a time interval is equal to the value of the average acceleration over the entire time interval: aa

if

if

tt

vva

• Let tttvvvv fifi ,0;,0

t

vva 0

atvv 0

One-dimensional Motion with Constant Acceleration

Displacement• When the velocity is increasing or decreasing uniformly with time, we can express the average velocity in any time interval as the arithmetic average of the initial velocity:

20 vv

v

• Now from the definition of displacement:

tatvvtvv

tvx

000

2

1

2

20 2

1attvx

v

Displacement• The area under the graph of v vs. t for any object is equal to the displacement x of the object.

20 2

1attvx

a

vvt

t

vva 00

tvv

tvx

2

0

a

vvvvt

vvtvx 0

00

2

1

2

One-dimensional Motion with Constant Acceleration

a

vvx

2

20

2 xavv 22

02or

Some examples

Examples of Constant Acceleration

• Example 2.5: Car chase

A trooper spotsa speeding carat 24.0 m/s

The troopersets off inchase at a=3.00 m/s2

a) How long does it take the trooper to overtake the speeding car? at time t.

t

vtxx

m/s) 0.24(m 0.240car

22

22

2troopertrooper

)m/s 50.1(

)m/s 00.3(2

12

1

t

t

tax

24.0 m

Examples of Constant Acceleration

A trooper spotsa speeding carat 24.0 m/s

The troopersets off inchase at a=3.00 m/s2

t

vtxx

m/s) 0.24(m 0.240car

22

22

2troopertrooper

)m/s 50.1(

)m/s 00.3(2

12

1

t

t

tax

troopercar xx Solve:

0m 0.24m/s) 0.24()m/s 50.1( 22 tt

s 9.16tb) At t the trooper’s speed is:

m/s 50.7s) 9.16)(m/s 00.3(0 2trooper0trooper tavv

• Example 2.5: Car chase (cont’d)

Examples of Constant Acceleration

• Example 2.7: Runaway length

m 71.50s) m/s)(1.00 5.71(2

1 20coasting attvx

braking20

2 2 xavv

m 572)m/s 47.4(00.2

m/s) (71.50

2 2

220

2

braking

a

vvx

m 644m 572 m 72brakingcoasting xx

Effect of gravity on freely falling objects

Freely Falling Objects

• When air resistance is negligible, all objects dropped under the influence of gravity near Earth’s surface fall toward Earth with the same constant acceleration. (Galileo Galilei, ~Year 1600)

• A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion.

• The magnitude of the free-fall acceleration is denoted by g. The value of g decreases with increasing altitude, and varies slightly with latitude, as well.

• At Earth’s surface, the value of g is approximately 9.80 m/s2.

Examples

Freely Falling Objects

• Example 2.9: Not a bad throw for a rookie!

y

xa) Find the time when the stone reaches its maximum height (v=0).

m/s 0.20)m/s 80.9( 20 tvatv

velocity:

The velocity is zero at the maximum height:

s. 04.2m/s 80.9

m/s 0.20

m/s 0.20)m/s 80.9(0

2

2

t

t

Freely Falling Objects

• Example 2.9: Not a bad throw for a rookie!

y

xb) Determine the stone’s maximum height.

22

200

)m/s 90.4(m/s) 0.20(

2

1

ttyy

attvyyy

y coordinate:

The velocity is zero at the maximum height:

s. 04.2m/s 80.9

m/s 0.202max

t

m. 4.20

)m/s 90.4(m/s) 0.20( 2max

2maxmax

tty

Freely Falling Objects

• Example 2.9: Not a bad throw for a rookie!

y

xc) Find the time the stone takes to return to its initial position (y=0) and its velocity.

22 )m/s 90.4(m/s) 0.20( tty y coordinate :

The y coordinate is zero at that moment :

s. 08.4,s 00.0

)m/s 90.4(m/s) 0.20(0 22

t

tt

The velocity at that moment is :

m/s. 0.20

m/s 0.20s) 08.4)(m/s 80.9( 2

v

Freely Falling Objects

• Example 2.9: Not a bad throw for a rookie!

y

xd) Find the time required for the stone to reach the ground (y=-50.0 m).

22 )m/s 90.4(m/s) 0.20( tty y coordinate :

The y coordinate is -50.0 m at that moment :

root). positive the(take s 83.5

)m/s 90.4(m/s) 0.20(m 0.50 22

t

tt

Freely Falling Objects

• Example 2.10: A rocket goes ballistic

y

x

a) Find the rocket’s velocity and position after 4.00 s.

y coordinate at t=4.0 s :

velocity at t=4.00 s :

m/s. 118

0.00)m/s 4.29( 20

tvatv

m. 235

)m/s 4.29(2

1m/s) 00.0(

2

1

22

20

tt

attvy t=4.00 s

Freely Falling Objects

• Example 2.10: A rocket goes ballistic

y

x

b) Find the maximum height the rocket attains.

y coordinate :

velocity is zero and = g :

s. 0.12

0.00m/s 118)m/s 80.9( 20

t

tvatv

m. 945

)m/s 80.9(2

1m/s) 118(m 235

2

1

22

200

tt

attvyy t=4.00 s

Freely Falling Objects

• Example 2.10: A rocket goes ballistic

y

x

c) Find the velocity of the rocket just prior to impact.

y coordinate = 0.00 :

velocity is zero and = g :

m/s. 136

m/s 118s) 9.25)(m/s 80.9( 20

vatv

s. 25.9tm 0.00

)m/s 80.9(2

1m/s) 118(m 235

2

1

22

200

tt

attvyy

t=4.00 s