motion in two and three dimensions - university of louisville

1Prof. Sergio B. MendesSummer 2018

Chapter 3 of Essential University Physics, Richard Wolfson, 3rd Edition

Motion in Two and Three Dimensions


Physical Quantities

VectorsScalars

• Temperature • Displacement

• Mass

• Pressure

• Velocity

• Acceleration

• Force• Volume

Magnitude (number & unit) Magnitude (number & unit) and Direction


Summing Two Vectors

𝑨𝑨

𝑩𝑩

𝑨𝑨 + 𝑩𝑩 = 𝑪𝑪

𝑩𝑩

𝑨𝑨𝑪𝑪

𝑨𝑨 + 𝑩𝑩 =

𝐶𝐶 = 𝐴𝐴2 + 𝐵𝐵2 + 2 𝐴𝐴 𝐵𝐵 𝑐𝑐𝑐𝑐𝑐𝑐 𝛼𝛼

𝛼𝛼

𝑨𝑨𝐴𝐴 = 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑐𝑐𝑜𝑜

𝑩𝑩𝐵𝐵 = 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑐𝑐𝑜𝑜

𝑪𝑪𝐶𝐶 = 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑐𝑐𝑜𝑜

𝑩𝑩

𝑨𝑨

𝑩𝑩 + 𝑨𝑨 = 𝑪𝑪


Summing Three Vectors

𝑨𝑨

𝑩𝑩

𝑨𝑨 + 𝑩𝑩 + 𝑪𝑪 = 𝑫𝑫

𝑪𝑪

𝑫𝑫

𝑨𝑨

𝑩𝑩

𝑪𝑪

𝑨𝑨 + 𝑩𝑩 + 𝑪𝑪

𝑨𝑨 + 𝑩𝑩𝑩𝑩 + 𝑪𝑪

𝑫𝑫

= 𝑫𝑫


Subtracting Two Vectors

𝑨𝑨

𝑨𝑨 − 𝑩𝑩 = 𝑪𝑪

𝑪𝑪

𝐶𝐶 = 𝐴𝐴2 + 𝐵𝐵2 − 2 𝐴𝐴 𝐵𝐵 𝑐𝑐𝑐𝑐𝑐𝑐 𝛼𝛼

𝛼𝛼

𝑩𝑩

𝑨𝑨

−𝑩𝑩

−𝑩𝑩


Multiplying a Vector by a Scalar𝑨𝑨 𝛽𝛽

𝛽𝛽 𝑨𝑨

𝑨𝑨

𝑨𝑨

𝛽𝛽 𝑨𝑨

If 𝛽𝛽 is positive

If 𝛽𝛽 is negative


Representing a Vector in a Cartesian Coordinate System: 2D

𝐴𝐴𝑥𝑥 = 𝐴𝐴 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃

𝐴𝐴𝑦𝑦 = 𝐴𝐴 𝑐𝑐𝑚𝑚𝑚𝑚 𝜃𝜃


Representing a Vector in terms of Unit Vectors: 2D

𝐴𝐴𝑥𝑥 = 𝐴𝐴 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃

𝐴𝐴𝑦𝑦 = 𝐴𝐴 𝑐𝑐𝑚𝑚𝑚𝑚 𝜃𝜃

𝑨𝑨 = 𝐴𝐴𝑥𝑥 �̂�𝒊 + 𝐴𝐴𝑦𝑦 �̂�𝒋


Representation of a Vector in terms of Unit Vectors: 3D

𝑨𝑨 = 𝐴𝐴𝑥𝑥 �̂�𝒊 + 𝐴𝐴𝑦𝑦 �̂�𝒋 + 𝐴𝐴𝑧𝑧 �𝒌𝒌


Example 3.1


Position Vector

𝒓𝒓 ≡ 𝑥𝑥 �̂�𝒊 + 𝑦𝑦 �̂�𝒋 + 𝑧𝑧 �𝒌𝒌𝒓𝒓

𝑥𝑥

𝑦𝑦

𝑧𝑧

𝒪𝒪


Displacement Vector∆𝒓𝒓 ≡ 𝒓𝒓𝟐𝟐 − 𝒓𝒓𝟏𝟏

𝑥𝑥

𝑦𝑦

𝑧𝑧

𝒪𝒪

𝒓𝒓𝟐𝟐𝒓𝒓𝟏𝟏

∆𝑥𝑥 = 𝑥𝑥2 − 𝑥𝑥1

∆𝑦𝑦 = 𝑦𝑦2 − 𝑦𝑦1

∆𝑧𝑧 = 𝑧𝑧2 − 𝑧𝑧1

∆𝒓𝒓 = 𝒓𝒓𝟐𝟐 − 𝒓𝒓𝟏𝟏

∆𝑥𝑥 �̂�𝒊 + ∆𝑦𝑦 �̂�𝒋 + ∆𝑧𝑧 �𝒌𝒌 = 𝑥𝑥2 − 𝑥𝑥1 �̂�𝒊 + 𝑦𝑦2 − 𝑦𝑦1 �̂�𝒋 + 𝑧𝑧2 − 𝑧𝑧1 �𝒌𝒌

= 𝒓𝒓𝟏𝟏 + ∆𝒓𝒓


Average Velocity Vector�𝒗𝒗 ≡

𝒓𝒓𝟐𝟐 − 𝒓𝒓𝟏𝟏𝑚𝑚2 − 𝑚𝑚1

𝑥𝑥

𝑦𝑦

𝑧𝑧

𝒪𝒪

𝒓𝒓𝟏𝟏

�̅�𝑣𝑥𝑥 =𝑥𝑥2 − 𝑥𝑥1𝑚𝑚2 − 𝑚𝑚1

=∆𝑥𝑥∆𝑚𝑚

�̅�𝑣𝑦𝑦 =𝑦𝑦2 − 𝑦𝑦1𝑚𝑚2 − 𝑚𝑚1

=∆𝑦𝑦∆𝑚𝑚

�̅�𝑣𝑧𝑧 =𝑧𝑧2 − 𝑧𝑧1𝑚𝑚2 − 𝑚𝑚1

=∆𝑧𝑧∆𝑚𝑚

∆𝒓𝒓 = 𝒓𝒓𝟐𝟐 − 𝒓𝒓𝟏𝟏 = �𝒗𝒗 𝑚𝑚2 − 𝑚𝑚1

𝒓𝒓𝟐𝟐 = 𝒓𝒓𝟏𝟏 + ∆𝒓𝒓

�̅�𝑣𝑥𝑥 �̂�𝒊 + �̅�𝑣𝑦𝑦 �̂�𝒋 + �̅�𝑣𝑧𝑧 �𝒌𝒌 = =∆𝒓𝒓∆𝑚𝑚


Instantaneous Velocity Vector

𝒗𝒗 ≡ lim∆𝑡𝑡→𝟎𝟎

∆𝒓𝒓∆𝑚𝑚

𝑣𝑣𝑥𝑥 = lim∆𝑡𝑡→𝟎𝟎

∆𝑥𝑥∆𝑚𝑚

=𝑚𝑚𝑥𝑥𝑚𝑚𝑚𝑚

𝑣𝑣𝑥𝑥 �̂�𝒊 + 𝑣𝑣𝑦𝑦 �̂�𝒋 + 𝑣𝑣𝑧𝑧 �𝒌𝒌 =

𝑣𝑣𝑦𝑦 = lim∆𝑡𝑡→𝟎𝟎

∆𝑦𝑦∆𝑚𝑚

=𝑚𝑚𝑦𝑦𝑚𝑚𝑚𝑚

𝑣𝑣𝑧𝑧 = lim∆𝑡𝑡→𝟎𝟎

∆𝑧𝑧∆𝑚𝑚

=𝑚𝑚𝑧𝑧𝑚𝑚𝑚𝑚


Average Acceleration Vector�𝒂𝒂 ≡

𝒗𝒗𝟐𝟐 − 𝒗𝒗𝟏𝟏𝑚𝑚2 − 𝑚𝑚1

𝑥𝑥

𝑦𝑦

𝑧𝑧

𝒪𝒪

𝒗𝒗𝟏𝟏

�𝑚𝑚𝑥𝑥 =𝑣𝑣2 − 𝑣𝑣1𝑚𝑚2 − 𝑚𝑚1

=∆𝑣𝑣∆𝑚𝑚

�𝑚𝑚𝑦𝑦 =𝑣𝑣2 − 𝑣𝑣1𝑚𝑚2 − 𝑚𝑚1

=∆𝑣𝑣∆𝑚𝑚

�𝑚𝑚𝑧𝑧 =𝑣𝑣2 − 𝑣𝑣1𝑚𝑚2 − 𝑚𝑚1

=∆𝑧𝑧∆𝑚𝑚

∆𝒗𝒗 = �𝒂𝒂 𝑚𝑚2 − 𝑚𝑚1

𝒗𝒗𝟐𝟐 = 𝒗𝒗𝟏𝟏 + ∆𝒗𝒗

�𝑚𝑚𝑥𝑥 �̂�𝒊 + �𝑚𝑚𝑦𝑦 �̂�𝒋 + �𝑚𝑚𝑧𝑧 �𝒌𝒌 = =∆𝒗𝒗∆𝑚𝑚


Instantaneous Acceleration Vector

𝒂𝒂 ≡ lim∆𝑡𝑡→𝟎𝟎

∆𝒗𝒗∆𝑚𝑚

𝑚𝑚𝑥𝑥 = lim∆𝑡𝑡→𝟎𝟎

∆𝑣𝑣𝑥𝑥∆𝑚𝑚

=𝑚𝑚𝑣𝑣𝑥𝑥𝑚𝑚𝑚𝑚

𝑚𝑚𝑥𝑥 �̂�𝒊 + 𝑚𝑚𝑦𝑦 �̂�𝒋 + 𝑚𝑚𝑧𝑧 �𝒌𝒌 =

𝑚𝑚𝑦𝑦 = lim∆𝑡𝑡→𝟎𝟎

∆𝑣𝑣𝑦𝑦∆𝑚𝑚

=𝑚𝑚𝑣𝑣𝑦𝑦𝑚𝑚𝑚𝑚

𝑚𝑚𝑧𝑧 = lim∆𝑡𝑡→𝟎𝟎

∆𝑣𝑣𝑧𝑧∆𝑚𝑚

=𝑚𝑚𝑣𝑣𝑧𝑧𝑚𝑚𝑚𝑚


A Few Observations


Same Direction 𝒗𝒗𝒐𝒐 & 𝒂𝒂

Magnitude of velocity changes but not the direction.


Opposite Direction 𝒗𝒗𝒐𝒐 & 𝒂𝒂

Magnitude of velocity changes and eventually the direction may be reversed.


Arbitrary Direction 𝒗𝒗𝒐𝒐 & 𝒂𝒂

Direction and magnitude of velocity change.


Velocity is always tangential to the trajectory

𝒗𝒗 ≡ lim∆𝑡𝑡→𝟎𝟎

∆𝒓𝒓∆𝑚𝑚 𝒗𝒗 ∥ ∆𝒓𝒓

A curved (non-straight) trajectory always requires acceleration !!


Relative Motion

∆𝒓𝒓𝑎𝑎𝑎𝑎𝑡𝑡, 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏

∆𝒓𝒓𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏, 𝑟𝑟𝑏𝑏𝑏𝑏𝑟𝑟

∆𝒓𝒓𝑎𝑎𝑎𝑎𝑡𝑡, 𝑟𝑟𝑏𝑏𝑏𝑏𝑟𝑟 = ∆𝒓𝒓𝑎𝑎𝑎𝑎𝑡𝑡, 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 + ∆𝒓𝒓𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏, 𝑟𝑟𝑏𝑏𝑏𝑏𝑟𝑟

∆𝒓𝒓𝑎𝑎𝑎𝑎𝑡𝑡, 𝑟𝑟𝑏𝑏𝑏𝑏𝑟𝑟

∆𝑚𝑚=∆𝒓𝒓𝑎𝑎𝑎𝑎𝑡𝑡, 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏

∆𝑚𝑚+∆𝒓𝒓𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏, 𝑟𝑟𝑏𝑏𝑏𝑏𝑟𝑟

∆𝑚𝑚lim∆𝑡𝑡→𝟎𝟎

𝒗𝒗𝑎𝑎𝑎𝑎𝑡𝑡, 𝑟𝑟𝑏𝑏𝑏𝑏𝑟𝑟 = 𝒗𝒗𝑎𝑎𝑎𝑎𝑡𝑡, 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 + 𝒗𝒗𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏, 𝑟𝑟𝑏𝑏𝑏𝑏𝑟𝑟


Example 3.2A jetliner flies at 960 km/h relative to the air in a wind blowing eastward at 190 km/h. It wants to go 1290 km straight northward. In what direction should the plane point to track northward ? How long will the trip take ?

𝒗𝒗𝑗𝑗, 𝑔𝑔 = 𝒗𝒗𝑗𝑗, 𝑎𝑎 + 𝒗𝒗𝑎𝑎, 𝑔𝑔

𝒗𝒗𝑗𝑗, 𝑔𝑔𝒗𝒗𝑗𝑗, 𝑎𝑎

𝒗𝒗𝑎𝑎, 𝑔𝑔

𝒗𝒗𝑎𝑎, 𝑔𝑔 = 𝑣𝑣𝑎𝑎, 𝑔𝑔 �̂�𝒊 = 190𝑘𝑘𝑚𝑚ℎ

�̂�𝒊

�̂�𝒊

�̂�𝒋

𝑣𝑣𝑗𝑗, 𝑎𝑎 = 960𝑘𝑘𝑚𝑚ℎ

𝒗𝒗𝑗𝑗, 𝑔𝑔 = 0 �̂�𝒊 + 𝑣𝑣𝑗𝑗, 𝑔𝑔 �̂�𝒋

𝒗𝒗𝑗𝑗, 𝑎𝑎 = 𝑣𝑣𝑗𝑗, 𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃 �̂�𝒊 + 𝑣𝑣𝑗𝑗, 𝑎𝑎 𝑐𝑐𝑚𝑚𝑚𝑚 𝜃𝜃 �̂�𝒋

What do we know ?

What do we want ?𝜃𝜃 = cos−1

−𝑣𝑣𝑎𝑎, 𝑔𝑔

𝑣𝑣𝑗𝑗, 𝑎𝑎= 101.4°

∆𝑚𝑚 =∆𝑦𝑦𝑣𝑣𝑗𝑗, 𝑔𝑔

=∆𝑦𝑦

𝑣𝑣𝑗𝑗, 𝑎𝑎 𝑐𝑐𝑚𝑚𝑚𝑚 𝜃𝜃= 1.4 h

∆𝑦𝑦 = 1290 𝑘𝑘𝑚𝑚


Constant Acceleration in 3D

𝒂𝒂 𝑚𝑚 = 𝒂𝒂

𝑚𝑚𝑥𝑥 𝑚𝑚 = 𝑚𝑚𝑥𝑥

𝑚𝑚𝑦𝑦 𝑚𝑚 = 𝑚𝑚𝑦𝑦

𝑚𝑚𝑧𝑧 𝑚𝑚 = 𝑚𝑚𝑧𝑧


From the definition of average velocity in 3D:

∆𝒗𝒗 = 𝒗𝒗 𝑚𝑚2 − 𝒗𝒗 𝑚𝑚1 = �𝒂𝒂 × 𝑚𝑚2 − 𝑚𝑚1

𝒗𝒗 𝑚𝑚 − 𝒗𝒗𝑏𝑏 = 𝒂𝒂 × 𝑚𝑚 − 0

�𝒂𝒂 = 𝒂𝒂

𝑚𝑚2 ≡ 𝑚𝑚

𝑚𝑚1 ≡ 0

𝒗𝒗 𝑚𝑚 = 𝒗𝒗𝑏𝑏 + 𝒂𝒂 𝑚𝑚

𝒗𝒗 𝑚𝑚1 ≡ 0 ≡ 𝒗𝒗𝑏𝑏



𝑣𝑣𝑥𝑥 𝑚𝑚 = 𝑣𝑣𝑏𝑏,𝑥𝑥 + 𝑚𝑚𝑥𝑥 𝑚𝑚

𝑣𝑣𝑦𝑦 𝑚𝑚 = 𝑣𝑣𝑏𝑏,𝑦𝑦 + 𝑚𝑚𝑦𝑦 𝑚𝑚

𝑣𝑣𝑧𝑧 𝑚𝑚 = 𝑣𝑣𝑏𝑏,𝑧𝑧 + 𝑚𝑚𝑧𝑧 𝑚𝑚


𝑣𝑣𝑥𝑥 = 𝑣𝑣𝑏𝑏,𝑥𝑥 + 𝑚𝑚𝑥𝑥 𝑚𝑚 𝑥𝑥 𝑚𝑚 = 𝑥𝑥𝑏𝑏 + 𝑣𝑣𝑏𝑏,𝑥𝑥 𝑚𝑚 +12𝑚𝑚𝑥𝑥 𝑚𝑚2

𝑣𝑣𝑦𝑦 = 𝑣𝑣𝑏𝑏,𝑦𝑦 + 𝑚𝑚𝑦𝑦 𝑚𝑚 𝑦𝑦 𝑚𝑚 = 𝑦𝑦𝑏𝑏 + 𝑣𝑣𝑏𝑏,𝑦𝑦 𝑚𝑚 +12𝑚𝑚𝑦𝑦 𝑚𝑚2

𝑣𝑣𝑧𝑧 = 𝑣𝑣𝑏𝑏,𝑧𝑧 + 𝑚𝑚𝑧𝑧 𝑚𝑚 𝑧𝑧 𝑚𝑚 = 𝑧𝑧𝑏𝑏 + 𝑣𝑣𝑏𝑏,𝑧𝑧 𝑚𝑚 +12𝑚𝑚𝑧𝑧 𝑚𝑚2

𝒓𝒓 𝑚𝑚 = 𝒓𝒓𝑏𝑏 + 𝒗𝒗𝑏𝑏𝑚𝑚 +12𝒂𝒂 𝑚𝑚𝟐𝟐

𝒓𝒓 𝑚𝑚 = 𝑥𝑥 𝑚𝑚 �̂�𝒊 + 𝑦𝑦 𝑚𝑚 �̂�𝒋 + 𝑧𝑧 𝑚𝑚 �𝒌𝒌

𝒗𝒗𝑏𝑏 = 𝑣𝑣𝑏𝑏,𝑥𝑥 �̂�𝒊 + 𝑣𝑣𝑏𝑏,𝑦𝑦 �̂�𝒋 + 𝑣𝑣𝑏𝑏,𝑧𝑧 �𝒌𝒌

𝒂𝒂 = 𝑚𝑚𝑥𝑥 �̂�𝒊 + 𝑚𝑚𝑥𝑥 �̂�𝒋 + 𝑚𝑚𝑥𝑥 �𝒌𝒌

𝒓𝒓𝑏𝑏 = 𝑥𝑥𝑏𝑏 �̂�𝒊 + 𝑦𝑦𝑏𝑏 �̂�𝒋 + 𝑧𝑧𝑏𝑏 �𝒌𝒌

�̂�𝒊

�̂�𝒋

�𝒌𝒌



𝒂𝒂 𝑚𝑚 = 𝒂𝒂

𝒓𝒓 𝑚𝑚 = 𝒓𝒓𝑏𝑏 + 𝒗𝒗𝑏𝑏 𝑚𝑚 +12𝒂𝒂 𝑚𝑚𝟐𝟐

Constant Acceleration in 3D, in Summary:


Bottom Line: We can study the motion in each Cartesian direction independently.


Example 3.3

𝑣𝑣𝑏𝑏,𝑥𝑥 = 7.3 𝑚𝑚/𝑐𝑐 𝑚𝑚𝑥𝑥 = 𝑚𝑚 𝑐𝑐𝑐𝑐𝑐𝑐 60°

𝑚𝑚𝑦𝑦 = 𝑚𝑚 𝑐𝑐𝑚𝑚𝑚𝑚 60°

𝑚𝑚 = 0.82 𝑚𝑚/𝑐𝑐2𝑣𝑣𝑏𝑏 = 7.3 𝑚𝑚/𝑐𝑐

𝑣𝑣𝑏𝑏,𝑦𝑦 = 0

You’re windsurfing at 7.3 m/s when a gust hits, accelerating your sailboard at 0.82 m/s2 at 60° to your original direction. If the gust last 8.7 s, what’s the board’s displacement during this time?

Δ𝑚𝑚 = 8.7 𝑐𝑐


𝑥𝑥 𝑚𝑚 − 𝑥𝑥𝑏𝑏 = 𝑣𝑣𝑏𝑏,𝑥𝑥 𝑚𝑚 +12𝑚𝑚𝑥𝑥 𝑚𝑚2

𝑣𝑣𝑏𝑏,𝑥𝑥 = 7.3 𝑚𝑚/𝑐𝑐

𝑚𝑚𝑥𝑥 = 𝑚𝑚 𝑐𝑐𝑐𝑐𝑐𝑐 60°

𝑚𝑚 = 0.82 𝑚𝑚/𝑐𝑐2

𝑚𝑚 = 8.7 𝑐𝑐

𝑥𝑥 𝑚𝑚 = 8.7 𝑐𝑐 − 𝑥𝑥𝑏𝑏 = 79.0 𝑚𝑚


𝑦𝑦 𝑚𝑚 − 𝑦𝑦𝑏𝑏 = 𝑣𝑣𝑏𝑏,𝑦𝑦 𝑚𝑚 +12𝑚𝑚𝑦𝑦 𝑚𝑚2


𝑚𝑚𝑦𝑦 = 𝑚𝑚 𝑐𝑐𝑚𝑚𝑚𝑚 60°

𝑚𝑚 = 0.82 𝑚𝑚/𝑐𝑐2

𝑚𝑚 = 8.7 𝑐𝑐

𝑦𝑦 𝑚𝑚 = 8.7 𝑐𝑐 − 𝑦𝑦𝑏𝑏 = 26.9 𝑚𝑚

∆𝑟𝑟 = 𝑥𝑥 − 𝑥𝑥𝑏𝑏 2 + 𝑦𝑦 − 𝑦𝑦𝑏𝑏 2 = 83 𝑚𝑚


Projectile Motion

Projectile Motion - PhET

from University of Colorado at Boulder

https://phet.colorado.edu/sims/html/projectile-motion/latest/projectile-motion_en.html


𝑣𝑣𝑥𝑥 = 𝑣𝑣𝑏𝑏,𝑥𝑥 𝑥𝑥 𝑚𝑚 = 𝑥𝑥𝑏𝑏 + 𝑣𝑣𝑏𝑏,𝑥𝑥 𝑚𝑚

𝑣𝑣𝑦𝑦 = 𝑣𝑣𝑏𝑏,𝑦𝑦 𝑦𝑦 𝑚𝑚 = 𝑦𝑦𝑏𝑏 + 𝑣𝑣𝑏𝑏,𝑦𝑦 𝑚𝑚

𝑦𝑦

𝑥𝑥

𝒂𝒂 𝑚𝑚 = −𝑚𝑚 �̂�𝒋

𝒗𝒗𝑏𝑏

𝒂𝒂 = −𝑚𝑚 �̂�𝒋

�̂�𝒋

𝒓𝒓𝑏𝑏

+12𝑚𝑚𝑥𝑥𝑚𝑚2+ 𝑚𝑚𝑥𝑥 𝑚𝑚

+ 𝑚𝑚𝑦𝑦 𝑚𝑚−𝑚𝑚 𝑚𝑚

𝑣𝑣𝑏𝑏,𝑥𝑥𝑣𝑣𝑏𝑏,𝑦𝑦

𝑥𝑥𝑏𝑏

𝑦𝑦𝑏𝑏

+12𝑚𝑚𝑦𝑦𝑚𝑚2−

12𝑚𝑚 𝑚𝑚2

𝑚𝑚𝑥𝑥 = 0

𝑚𝑚𝑦𝑦 = −𝑚𝑚


𝑥𝑥 𝑚𝑚𝑔𝑔𝑟𝑟 − 𝑥𝑥𝑏𝑏 = 𝑣𝑣𝑏𝑏,𝑥𝑥 𝑚𝑚𝑔𝑔𝑟𝑟 = ? ?

𝑦𝑦 𝑚𝑚𝑔𝑔𝑟𝑟 = 𝑦𝑦𝑏𝑏 + 𝑣𝑣𝑏𝑏,𝑦𝑦𝑚𝑚𝑔𝑔𝑟𝑟 −12𝑚𝑚 𝑚𝑚𝑔𝑔𝑟𝑟2


𝑦𝑦 𝑚𝑚𝑔𝑔𝑟𝑟 − 𝑦𝑦𝑏𝑏 = −1.7 𝑚𝑚

𝑣𝑣𝑏𝑏,𝑥𝑥 = 31 𝑚𝑚/𝑐𝑐

𝑚𝑚𝑔𝑔𝑟𝑟 = ? ?

Example 3.4

𝑚𝑚𝑔𝑔𝑟𝑟 = −2𝑦𝑦 𝑚𝑚𝑔𝑔𝑟𝑟 − 𝑦𝑦𝑏𝑏

𝑚𝑚= 0.589 𝑐𝑐𝑥𝑥 𝑚𝑚𝑔𝑔𝑟𝑟 − 𝑥𝑥𝑏𝑏 = 𝑣𝑣𝑏𝑏,𝑥𝑥 𝑚𝑚𝑔𝑔𝑟𝑟 = 18 𝑚𝑚


𝑥𝑥 𝑚𝑚 = 𝑥𝑥𝑏𝑏 + 𝑣𝑣𝑏𝑏,𝑥𝑥 𝑚𝑚

𝑦𝑦 𝑚𝑚 = 𝑦𝑦𝑏𝑏 + 𝑣𝑣𝑏𝑏,𝑦𝑦 𝑚𝑚 −12𝑚𝑚 𝑚𝑚2

𝑦𝑦

𝑥𝑥

𝒗𝒗𝑏𝑏

𝒂𝒂 = −𝑚𝑚 �̂�𝒋

�̂�𝒋

𝒓𝒓𝑏𝑏

𝑣𝑣𝑏𝑏,𝑥𝑥𝑣𝑣𝑏𝑏,𝑦𝑦

𝑥𝑥𝑏𝑏

𝑦𝑦𝑏𝑏

What kind of trajectory ?

𝑦𝑦 = 𝑦𝑦𝑏𝑏 + 𝑣𝑣𝑏𝑏,𝑦𝑦𝑥𝑥 − 𝑥𝑥𝑏𝑏𝑣𝑣𝑏𝑏,𝑥𝑥

−12𝑚𝑚

𝑥𝑥 − 𝑥𝑥𝑏𝑏𝑣𝑣𝑏𝑏,𝑥𝑥

2

Parabola !!


Range of a Projectile

𝑦𝑦 = 𝑦𝑦𝑏𝑏 + 𝑣𝑣𝑏𝑏,𝑦𝑦𝑥𝑥 − 𝑥𝑥𝑏𝑏𝑣𝑣𝑏𝑏,𝑥𝑥

−12𝑚𝑚

𝑥𝑥 − 𝑥𝑥𝑏𝑏𝑣𝑣𝑏𝑏,𝑥𝑥

2

𝑦𝑦 = 𝑦𝑦𝑏𝑏

𝑥𝑥 − 𝑥𝑥𝑏𝑏 =2 𝑣𝑣𝑏𝑏,𝑥𝑥 𝑣𝑣𝑏𝑏,𝑦𝑦

𝑚𝑚=𝑣𝑣𝑏𝑏2 𝑐𝑐𝑚𝑚𝑚𝑚 2 𝜃𝜃𝑏𝑏

𝑚𝑚

𝑦𝑦 = 𝑦𝑦𝑏𝑏


𝑥𝑥 − 𝑥𝑥𝑏𝑏 =𝑣𝑣𝑏𝑏2 𝑐𝑐𝑚𝑚𝑚𝑚 2 𝜃𝜃𝑏𝑏

𝑚𝑚

𝑚𝑚 =2 𝑣𝑣𝑏𝑏 𝑐𝑐𝑚𝑚𝑚𝑚 𝜃𝜃𝑏𝑏

𝑚𝑚

Horizontal range:

Time to return to the same height:


Uniform Circular Velocity:circular motion with a constant magnitude of the velocity


∆𝑟𝑟𝑟𝑟

=∆𝑣𝑣𝑣𝑣

𝑣𝑣1 = 𝑣𝑣2𝑟𝑟1 = 𝑟𝑟2𝑟𝑟1 = 𝑟𝑟2𝑣𝑣1 = 𝑣𝑣2

𝑚𝑚 =𝑣𝑣2

𝑟𝑟velocity and acceleration are continuously changing (as their

directions are changing), although their magnitudes are constant


Summary You learned to express motion quantities as vectors in one, two,

and three dimensions.

You learned that acceleration can change the velocity’s magnitude, direction, or both.

You can describe motion quantitatively when acceleration is constant.

You became familiar with projectile motion under the influence of gravity near Earth’s surface.

You became familiar with uniform circular motion.

motion in two and three dimensions - university of louisville

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