CURVILINE
AR MOTION
Curvilinear Motion
Curvilinear Motion
– is a motion that occurs when
a particle travels along a curved
path.
Curvilinear Motion in Space
Curvilinear Motion in Space
Motion of car along a
curved road.
Curvilinear Motion in Space
Motion of car along a
curved road.
Motion of cable car
along a steel cable.
Curvilinear Motion in Space
Curvilinear Motion in Space
Motion of roller
coaster along its
track.
Curvilinear Motion in Space
Motion of roller
coaster along its
track.
Motion of fighter jets
during national
parade.
Velocity in Curvilinear Motion
Velocity in Curvilinear Motion
• Position Vector
𝒓 = 𝑥𝒊 + 𝑦𝒋
𝑣𝑦 𝒋
• Velocity Vector
𝑑𝒓
𝑑𝑥
𝑑𝑦
𝒗 = = 𝒊 + 𝒋 = 𝑣𝑥 𝒊 +
𝑑𝑡
𝑑𝑡
𝑑𝑡
Velocity in Curvilinear Motion
• Position Vector
𝒓 = 𝑥𝒊 + 𝑦𝒋
𝑣𝑦 𝒋
• Velocity Vector
𝑑𝒓
𝑑𝑥
𝑑𝑦
𝒗 = = 𝒊 + 𝒋 = 𝑣𝑥 𝒊 +
𝑑𝑡
𝑑𝑡
𝑑𝑡
Velocity in Curvilinear Motion
• Position Vector
𝒓 = 𝑥𝒊 + 𝑦𝒋
𝑣𝑦 𝒋
• Velocity Vector
𝑑𝒓
𝑑𝑥
𝑑𝑦
𝒗 = = 𝒊 + 𝒋 = 𝑣𝑥 𝒊 +
𝑑𝑡
• Magnitude of 𝒗/Speed
𝒗 = 𝒗∙𝒗=
𝑣𝑥2
+ 𝑣𝑦2
=
𝑑𝑠
𝑑𝑡
𝑑𝑡
𝑑𝑡
Velocity in Curvilinear Motion
• Position Vector
𝒓 = 𝑥𝒊 + 𝑦𝒋
𝑣𝑦 𝒋
• Velocity Vector
𝑑𝒓
𝑑𝑥
𝑑𝑦
𝒗 = = 𝒊 + 𝒋 = 𝑣𝑥 𝒊 +
𝑑𝑡
• Magnitude of 𝒗/Speed
𝒗 = 𝒗∙𝒗=
𝑣𝑥2
+ 𝑣𝑦2
=
𝑑𝑡
𝑑𝑡
• Direction
𝑑𝑠
𝑑𝑡
tan 𝜏 =
𝑣𝑦
𝑣𝑥
Exampl
e:
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑑𝑡
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑑𝑡
At 𝑡 = 1
𝑣𝑦 = 5
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑑𝑡
At 𝑡 = 1
𝑣𝑦 = 5
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑑𝑡
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 = −𝒊 + 5𝒋
At 𝑡 = 1
𝑣𝑦 = 5
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑑𝑡
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 = −𝒊 + 5𝒋
At 𝑡 = 1
𝑣𝑦 = 5
– velocity vector
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑑𝑡
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 = −𝒊 + 5𝒋
𝒗 =
𝑣𝑥2 + 𝑣𝑦2
At 𝑡 = 1
𝑣𝑦 = 5
– velocity vector
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑑𝑡
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 = −𝒊 + 5𝒋
𝒗 =
𝑣𝑥2 + 𝑣𝑦2 =
(−1)2 +(5)2
At 𝑡 = 1
𝑣𝑦 = 5
– velocity vector
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑑𝑡
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 = −𝒊 + 5𝒋
𝒗 =
𝑣𝑥2 + 𝑣𝑦2 =
(−1)2 +(5)2 = 26
At 𝑡 = 1
𝑣𝑦 = 5
– velocity vector
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑑𝑡
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 = −𝒊 + 5𝒋
𝒗 =
𝑣𝑥2 + 𝑣𝑦2 =
At 𝑡 = 1
𝑣𝑦 = 5
– velocity vector
(−1)2 +(5)2 = 26 – magnitude/speed
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑑𝑡
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 = −𝒊 + 5𝒋
𝒗 =
tan 𝜏 =
𝑣𝑥2 + 𝑣𝑦2 =
𝑣𝑦
𝑣𝑥
At 𝑡 = 1
𝑣𝑦 = 5
– velocity vector
(−1)2 +(5)2 = 26 – magnitude/speed
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑𝑡
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 = −𝒊 + 5𝒋
𝒗 =
tan 𝜏 =
𝑣𝑥2 + 𝑣𝑦2 =
𝑣𝑦
𝑣𝑥
=
5
−1
At 𝑡 = 1
𝑣𝑦 = 5
– velocity vector
(−1)2 +(5)2 = 26 – magnitude/speed
= −5
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑𝑡
– velocity vector
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 = −𝒊 + 5𝒋
𝒗 =
tan 𝜏 =
𝑣𝑥2 + 𝑣𝑦2 =
𝑣𝑦
𝑣𝑥
=
5
−1
At 𝑡 = 1
𝑣𝑦 = 5
(−1)2 +(5)2 = 26 – magnitude/speed
= −5
𝜏 = tan−1 −5
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑𝑡
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 = −𝒊 + 5𝒋
𝒗 =
tan 𝜏 =
𝑣𝑥2 + 𝑣𝑦2 =
𝑣𝑦
𝑣𝑥
=
5
−1
At 𝑡 = 1
𝑣𝑦 = 5
– velocity vector
(−1)2 +(5)2 = 26 – magnitude/speed
= −5
𝜏 = tan−1 −5 = −78°41′
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑𝑡
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 = −𝒊 + 5𝒋
𝒗 =
tan 𝜏 =
𝑣𝑥2 + 𝑣𝑦2 =
𝑣𝑦
𝑣𝑥
=
5
−1
– velocity vector
(−1)2 +(5)2 = 26 – magnitude/speed
= −5
𝜏 = 180° − 78°41′
At 𝑡 = 1
𝑣𝑦 = 5
𝜏 = tan−1 −5 = −78°41′
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑𝑡
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 = −𝒊 + 5𝒋
𝒗 =
tan 𝜏 =
𝑣𝑥2 + 𝑣𝑦2 =
𝑣𝑦
𝑣𝑥
=
5
−1
At 𝑡 = 1
𝑣𝑦 = 5
– velocity vector
(−1)2 +(5)2 = 26 – magnitude/speed
= −5
𝜏 = 180° − 78°41′ = 101°19′
𝜏 = tan−1 −5 = −78°41′
Exampl
e: Find the
magnitude and direction of velocity at time t,
given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑𝑡
𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 = −𝒊 + 5𝒋
𝒗 =
tan 𝜏 =
𝑣𝑥2 + 𝑣𝑦2 =
𝑣𝑦
𝑣𝑥
=
5
−1
At 𝑡 = 1
𝑣𝑦 = 5
– velocity vector
(−1)2 +(5)2 = 26 – magnitude/speed
= −5
𝜏 = 180° − 78°41′ = 101°19′
𝜏 = tan−1 −5 = −78°41′
– direction
Acceleration in Curvilinear
Motion
Acceleration in Curvilinear
Motion
• Acceleration Vector
𝒂=
𝑑𝒗
𝑑𝑡
=
𝑑2 𝒓
𝑑𝑡 2
=
𝑑2 𝑥
𝒊
𝑑𝑡 2
+
𝑑2 𝑦
𝒋
𝑑𝑡 2
= 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋
• Magnitude of of 𝒂
𝒂 = 𝒂∙𝒂=
𝑎𝑥2 + 𝑎𝑦2
• Direction
𝑎𝑦
tan ∅ =
𝑎𝑥
Acceleration in Curvilinear
Motion
• Acceleration Vector
𝒂=
𝑑𝒗
𝑑𝑡
=
𝑑2 𝒓
𝑑𝑡 2
=
𝑑2 𝑥
𝒊
𝑑𝑡 2
+
𝑑2 𝑦
𝒋
𝑑𝑡 2
= 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋
• Magnitude of of 𝒂
𝒂 = 𝒂∙𝒂=
𝑎𝑥2 + 𝑎𝑦2
• Direction
𝑎𝑦
tan ∅ =
𝑎𝑥
Acceleration in Curvilinear
Motion
• Acceleration Vector
𝒂=
𝑑𝒗
𝑑𝑡
=
𝑑2 𝒓
𝑑𝑡 2
=
𝑑2 𝑥
𝒊
𝑑𝑡 2
+
𝑑2 𝑦
𝒋
𝑑𝑡 2
= 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋
• Magnitude of of 𝒂
𝒂 = 𝒂∙𝒂=
𝑎𝑥2 + 𝑎𝑦2
• Direction
𝑎𝑦
tan ∅ =
𝑎𝑥
Exampl
e:
Exampl
e: Find the magnitude and direction of acceleration at time
t, given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
Exampl
e: Find the magnitude and direction of acceleration at time
t, given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑2𝑥
𝑑𝑡 2
= 𝑎𝑥 = 0
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
At 𝑡 = 1
= 𝑎𝑦 = 12𝑡
𝑎𝑦 = 12
𝑑𝑡
𝑑2𝑦
𝑑𝑡 2
Exampl
e: Find the magnitude and direction of acceleration at time
t, given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑2𝑥
𝑑𝑡 2
= 𝑎𝑥 = 0
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
At 𝑡 = 1
= 𝑎𝑦 = 12𝑡
𝑎𝑦 = 12
𝑑𝑡
𝑑2𝑦
𝑑𝑡 2
Exampl
e: Find the magnitude and direction of acceleration at time
t, given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑2𝑥
𝑑𝑡 2
= 𝑎𝑥 = 0
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
At 𝑡 = 1
= 𝑎𝑦 = 12𝑡
𝑎𝑦 = 12
𝑑𝑡
𝑑2𝑦
𝑑𝑡 2
Exampl
e: Find the magnitude and direction of acceleration at time
t, given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑2𝑥
𝑑𝑡 2
= 𝑎𝑥 = 0
𝒂 = 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
At 𝑡 = 1
= 𝑎𝑦 = 12𝑡
𝑎𝑦 = 12
𝑑𝑡
𝑑2𝑦
𝑑𝑡 2
Exampl
e: Find the magnitude and direction of acceleration at time
t, given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑2𝑥
𝑑𝑡 2
= 𝑎𝑥 = 0
𝒂 = 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋 = 12𝒋
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
At 𝑡 = 1
= 𝑎𝑦 = 12𝑡
𝑎𝑦 = 12
𝑑𝑡
𝑑2𝑦
𝑑𝑡 2
Exampl
e: Find the magnitude and direction of acceleration at time
t, given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑2𝑥
𝑑𝑡 2
= 𝑎𝑥 = 0
𝒂 = 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋 = 12𝒋
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
At 𝑡 = 1
= 𝑎𝑦 = 12𝑡
𝑎𝑦 = 12
𝑑𝑡
𝑑2𝑦
𝑑𝑡 2
– acceleration vector
Exampl
e: Find the magnitude and direction of acceleration at time
t, given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑2𝑥
𝑑𝑡 2
= 𝑎𝑥 = 0
𝒂 = 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋 = 12𝒋
𝒂 =
𝑎𝑥2 + 𝑎𝑦2
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
At 𝑡 = 1
= 𝑎𝑦 = 12𝑡
𝑎𝑦 = 12
𝑑𝑡
𝑑2𝑦
𝑑𝑡 2
– acceleration vector
Exampl
e: Find the magnitude and direction of acceleration at time
t, given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑2𝑥
𝑑𝑡 2
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
At 𝑡 = 1
= 𝑎𝑦 = 12𝑡
𝑎𝑦 = 12
𝑑𝑡
𝑑2𝑦
𝑑𝑡 2
= 𝑎𝑥 = 0
𝒂 = 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋 = 12𝒋
𝒂 =
𝑎𝑥2 + 𝑎𝑦2 =
(0)2 +(12)2 = 12
– acceleration vector
Exampl
e: Find the magnitude and direction of acceleration at time
t, given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑2𝑥
𝑑𝑡 2
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
At 𝑡 = 1
= 𝑎𝑦 = 12𝑡
𝑎𝑦 = 12
𝑑𝑡
𝑑2𝑦
𝑑𝑡 2
= 𝑎𝑥 = 0
𝒂 = 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋 = 12𝒋
𝒂 =
𝑎𝑥2 + 𝑎𝑦2 =
(0)2 +(12)2 = 12
– acceleration vector
– magnitude
Exampl
e: Find the magnitude and direction of acceleration at time
t, given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑2𝑥
𝑑𝑡 2
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
At 𝑡 = 1
= 𝑎𝑦 = 12𝑡
𝑎𝑦 = 12
𝑑𝑡
𝑑2𝑦
𝑑𝑡 2
= 𝑎𝑥 = 0
𝒂 = 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋 = 12𝒋
𝒂 =
tan ∅ =
𝑎𝑥2 + 𝑎𝑦2 =
𝑎𝑦
𝑎𝑥
(0)2 +(12)2 = 12
– acceleration vector
– magnitude
Exampl
e: Find the magnitude and direction of acceleration at time
t, given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑2𝑥
𝑑𝑡 2
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
At 𝑡 = 1
= 𝑎𝑦 = 12𝑡
𝑎𝑦 = 12
𝑑𝑡
𝑑2𝑦
𝑑𝑡 2
= 𝑎𝑥 = 0
– acceleration vector
𝒂 = 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋 = 12𝒋
𝒂 =
tan ∅ =
𝑎𝑥2 + 𝑎𝑦2 =
𝑎𝑦
𝑎𝑥
(0)2 +(12)2 = 12
– magnitude
∅ = 90°
– direction
Exampl
e: Find the magnitude and direction of acceleration at time
t, given 𝑥 = 2 − 𝑡, 𝑦 = 2𝑡 3 − 𝑡; at 𝑡 = 1.
𝑥 =2−𝑡
𝑑𝑥
= 𝑣𝑥 = −1
𝑑𝑡
𝑑2𝑥
𝑑𝑡 2
𝑦 = 2𝑡 3 − 𝑡
𝑑𝑦
= 𝑣𝑦 = 6𝑡 2 − 1
At 𝑡 = 1
= 𝑎𝑦 = 12𝑡
𝑎𝑦 = 12
𝑑𝑡
𝑑2𝑦
𝑑𝑡 2
= 𝑎𝑥 = 0
– acceleration vector
𝒂 = 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋 = 12𝒋
𝒂 =
tan ∅ =
𝑎𝑥2 + 𝑎𝑦2 =
𝑎𝑦
𝑎𝑥
(0)2 +(12)2 = 12
∅ = 90°
– magnitude
Tangential and Normal
Components of Acceleration
Tangential and Normal
Components of Acceleration
• Tangent
𝑎𝑡 =
𝑑2 𝑠
𝑑𝑡 2
Tangential and Normal
Components of Acceleration
• Tangent
𝑎𝑡 =
𝑑2 𝑠
𝑑𝑡 2
• Normal
𝑎𝑛 =
𝒗2
𝑅
or
𝑎𝑛2 = 𝒂
2
− 𝑎𝑡2
Exampl
e:
Exampl
e: Find the
magnitudes of the tangential and normal
components of acceleration for the motion 𝑥 = 𝑒 𝑡 cos 𝑡, 𝑦 =
𝑒 𝑡 sin 𝑡 at any time 𝑡.
Exampl
e: Find the
magnitudes of the tangential and normal
components of acceleration for the motion 𝑥 = 𝑒 𝑡 cos 𝑡, 𝑦 =
𝑒 𝑡𝑥sin
time 𝑡.
= 𝑒𝑡𝑡 at
cosany
𝑡
𝑦 = 𝑒 𝑡 sin 𝑡
𝑑𝑥
= 𝑣𝑥 = 𝑒 𝑡 cos 𝑡 − sin 𝑡
𝑑𝑡
𝑑2𝑥
= 𝑎𝑥 = −2𝑒 𝑡 sin 𝑡
2
𝑑𝑡
𝑑𝑦
= 𝑣𝑦 = 𝑒 𝑡 sin 𝑡 +
𝑑𝑡
𝑑2𝑦
= 𝑎𝑦 = 2𝑒 𝑡 cos 𝑡
2
𝑑𝑡
cos 𝑡
Exampl
e: Find the
magnitudes of the tangential and normal
components of acceleration for the motion 𝑥 = 𝑒 𝑡 cos 𝑡, 𝑦 =
𝑒 𝑡𝑥sin
time 𝑡.
= 𝑒𝑡𝑡 at
cosany
𝑡
𝑦 = 𝑒 𝑡 sin 𝑡
𝑑𝑥
= 𝑣𝑥 = 𝑒 𝑡 cos 𝑡 − sin 𝑡
𝑑𝑡
𝑑2𝑥
= 𝑎𝑥 = −2𝑒 𝑡 sin 𝑡
2
𝑑𝑡
𝑑𝑠
𝑑𝑡
= 𝒗 =
𝑣𝑥2 + 𝑣𝑦2 =
𝑑𝑦
= 𝑣𝑦 = 𝑒 𝑡 sin 𝑡 +
𝑑𝑡
𝑑2𝑦
= 𝑎𝑦 = 2𝑒 𝑡 cos 𝑡
2
𝑑𝑡
𝑒𝑡 cos 𝑡 − sin 𝑡
2
cos 𝑡
+ 𝑒𝑡 sin 𝑡 + cos 𝑡
2
= 2𝑒 𝑡
Exampl
e: Find the
magnitudes of the tangential and normal
components of acceleration for the motion 𝑥 = 𝑒 𝑡 cos 𝑡, 𝑦 =
𝑒 𝑡𝑥sin
time 𝑡.
= 𝑒𝑡𝑡 at
cosany
𝑡
𝑦 = 𝑒 𝑡 sin 𝑡
𝑑𝑥
= 𝑣𝑥 = 𝑒 𝑡 cos 𝑡 − sin 𝑡
𝑑𝑡
𝑑2𝑥
= 𝑎𝑥 = −2𝑒 𝑡 sin 𝑡
2
𝑑𝑡
𝑑𝑠
= 𝒗
𝑑𝑡
𝑑2 𝑠
=
𝑑𝑡 2
=
2𝑒 𝑡
𝑣𝑥2 + 𝑣𝑦2 =
𝑑𝑦
= 𝑣𝑦 = 𝑒 𝑡 sin 𝑡 +
𝑑𝑡
𝑑2𝑦
= 𝑎𝑦 = 2𝑒 𝑡 cos 𝑡
2
𝑑𝑡
𝑒𝑡 cos 𝑡 − sin 𝑡
2
cos 𝑡
+ 𝑒𝑡 sin 𝑡 + cos 𝑡
2
= 2𝑒 𝑡
Exampl
e: Find the
magnitudes of the tangential and normal
components of acceleration for the motion 𝑥 = 𝑒 𝑡 cos 𝑡, 𝑦 =
𝑒 𝑡𝑥sin
time 𝑡.
= 𝑒𝑡𝑡 at
cosany
𝑡
𝑦 = 𝑒 𝑡 sin 𝑡
𝑑𝑥
= 𝑣𝑥 = 𝑒 𝑡 cos 𝑡 − sin 𝑡
𝑑𝑡
𝑑2𝑥
= 𝑎𝑥 = −2𝑒 𝑡 sin 𝑡
2
𝑑𝑡
𝑑𝑠
= 𝒗
𝑑𝑡
𝑑2 𝑠
=
𝑑𝑡 2
𝑎𝑡 =
𝑣𝑥2 + 𝑣𝑦2 =
=
𝑑𝑦
= 𝑣𝑦 = 𝑒 𝑡 sin 𝑡 +
𝑑𝑡
𝑑2𝑦
= 𝑎𝑦 = 2𝑒 𝑡 cos 𝑡
2
𝑑𝑡
𝑒𝑡 cos 𝑡 − sin 𝑡
2
cos 𝑡
+ 𝑒𝑡 sin 𝑡 + cos 𝑡
2
= 2𝑒 𝑡
2𝑒 𝑡
𝑑2𝑠
𝑑𝑡 2
= 2𝑒 𝑡
– magnitudes of the tangential
Exampl
e: Find the
magnitudes of the tangential and normal
components of acceleration for the motion 𝑥 = 𝑒 𝑡 cos 𝑡, 𝑦 =
𝑒 𝑡𝑥sin
time 𝑡.
= 𝑒𝑡𝑡 at
cosany
𝑡
𝑦 = 𝑒 𝑡 sin 𝑡
𝑑𝑥
= 𝑣𝑥 = 𝑒 𝑡 cos 𝑡 − sin 𝑡
𝑑𝑡
𝑑2𝑥
= 𝑎𝑥 = −2𝑒 𝑡 sin 𝑡
2
𝑑𝑡
𝑑2𝑠
𝑎𝑡 = 2 = 2𝑒 𝑡
𝑑𝑡
𝑑𝑦
= 𝑣𝑦 = 𝑒 𝑡 sin 𝑡 +
𝑑𝑡
𝑑2𝑦
= 𝑎𝑦 = 2𝑒 𝑡 cos 𝑡
2
𝑑𝑡
cos 𝑡
Exampl
e: Find the
magnitudes of the tangential and normal
components of acceleration for the motion 𝑥 = 𝑒 𝑡 cos 𝑡, 𝑦 =
𝑒 𝑡𝑥sin
time 𝑡.
= 𝑒𝑡𝑡 at
cosany
𝑡
𝑦 = 𝑒 𝑡 sin 𝑡
𝑑𝑥
= 𝑣𝑥 = 𝑒 𝑡 cos 𝑡 − sin 𝑡
𝑑𝑡
𝑑2𝑥
= 𝑎𝑥 = −2𝑒 𝑡 sin 𝑡
2
𝑑𝑡
𝑑2𝑠
𝑎𝑡 = 2 = 2𝑒 𝑡
𝑑𝑡
𝒂 =
𝑎𝑥2 + 𝑎𝑦2 =
−2𝑒 𝑡 sin 𝑡
𝑑𝑦
= 𝑣𝑦 = 𝑒 𝑡 sin 𝑡 +
𝑑𝑡
𝑑2𝑦
= 𝑎𝑦 = 2𝑒 𝑡 cos 𝑡
2
𝑑𝑡
2
+ 2𝑒 𝑡 cos 𝑡
2
cos 𝑡
= 2𝑒 𝑡
Exampl
e: Find the
magnitudes of the tangential and normal
components of acceleration for the motion 𝑥 = 𝑒 𝑡 cos 𝑡, 𝑦 =
𝑒 𝑡𝑥sin
time 𝑡.
= 𝑒𝑡𝑡 at
cosany
𝑡
𝑦 = 𝑒 𝑡 sin 𝑡
𝑑𝑥
= 𝑣𝑥 = 𝑒 𝑡 cos 𝑡 − sin 𝑡
𝑑𝑡
𝑑2𝑥
= 𝑎𝑥 = −2𝑒 𝑡 sin 𝑡
2
𝑑𝑡
𝑑2𝑠
𝑎𝑡 = 2 = 2𝑒 𝑡
𝑑𝑡
𝒂 =
𝑎𝑛2
𝑎𝑥2 + 𝑎𝑦2 =
2
= 𝒂 −
𝑎𝑡2
𝑑𝑦
= 𝑣𝑦 = 𝑒 𝑡 sin 𝑡 +
𝑑𝑡
𝑑2𝑦
= 𝑎𝑦 = 2𝑒 𝑡 cos 𝑡
2
𝑑𝑡
−2𝑒 𝑡 sin 𝑡
2
+ 2𝑒 𝑡 cos 𝑡
𝑎𝑛 =
2
𝒂 −
2
𝑎𝑡2
cos 𝑡
= 2𝑒 𝑡
=
2𝑒 𝑡 2
−
2𝑒 𝑡
2
Exampl
e: Find the
magnitudes of the tangential and normal
components of acceleration for the motion 𝑥 = 𝑒 𝑡 cos 𝑡, 𝑦 =
𝑒 𝑡𝑥sin
time 𝑡.
= 𝑒𝑡𝑡 at
cosany
𝑡
𝑦 = 𝑒 𝑡 sin 𝑡
𝑑𝑥
= 𝑣𝑥 = 𝑒 𝑡 cos 𝑡 − sin 𝑡
𝑑𝑡
𝑑2𝑥
= 𝑎𝑥 = −2𝑒 𝑡 sin 𝑡
2
𝑑𝑡
𝑑2𝑠
𝑎𝑡 = 2 = 2𝑒 𝑡
𝑑𝑡
𝒂 =
𝑎𝑛2
𝑎𝑥2 + 𝑎𝑦2 =
2
= 𝒂 −
𝑎𝑡2
𝑎𝑛 = 2𝑒 𝑡
𝑑𝑦
= 𝑣𝑦 = 𝑒 𝑡 sin 𝑡 +
𝑑𝑡
𝑑2𝑦
= 𝑎𝑦 = 2𝑒 𝑡 cos 𝑡
2
𝑑𝑡
−2𝑒 𝑡 sin 𝑡
2
+ 2𝑒 𝑡 cos 𝑡
𝑎𝑛 =
2
𝒂 −
2
𝑎𝑡2
cos 𝑡
= 2𝑒 𝑡
=
2𝑒 𝑡 2
−
– magnitudes of the normal
2𝑒 𝑡
2
Assessmen
t:
1.
2.
Find the magnitude and direction of velocity
and acceleration of time 𝑡 , given 𝑥 =
𝜋
cos 3𝑡, 𝑦 = sin 𝑡 at 𝑡 = .
4
Find the magnitude and direction of velocity
and acceleration for the motion 𝑥 = 3𝑡 and 𝑦 =
9𝑡 − 3𝑡 2 at 𝑡 = 2.