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 π = π₯π + π¦π π£π¦ π • Velocity Vector ππ ππ₯ ππ¦ π = = π + π = π£π₯ π + ππ‘ ππ‘ ππ‘ Velocity in Curvilinear Motion • Position Vector π = π₯π + π¦π π£π¦ π • Velocity Vector ππ ππ₯ ππ¦ π = = π + π = π£π₯ π + ππ‘ ππ‘ ππ‘ Velocity in Curvilinear Motion • Position Vector π = π₯π + π¦π π£π¦ π • Velocity Vector ππ ππ₯ ππ¦ π = = π + π = π£π₯ π + ππ‘ • Magnitude of π/Speed π = πβπ= π£π₯2 + π£π¦2 = ππ ππ‘ ππ‘ ππ‘ Velocity in Curvilinear Motion • Position Vector π = π₯π + π¦π π£π¦ π • Velocity Vector ππ ππ₯ ππ¦ π = = π + π = π£π₯ π + ππ‘ • Magnitude of π/Speed π = πβπ= π£π₯2 + π£π¦2 = ππ‘ ππ‘ • Direction ππ ππ‘ tan π = π£π¦ π£π₯ Exampl e: Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 ππ‘ Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 ππ‘ At π‘ = 1 π£π¦ = 5 Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π = π£π₯ π + π£π¦ π π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 ππ‘ At π‘ = 1 π£π¦ = 5 Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 ππ‘ π = π£π₯ π + π£π¦ π = −π + 5π At π‘ = 1 π£π¦ = 5 Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 ππ‘ π = π£π₯ π + π£π¦ π = −π + 5π At π‘ = 1 π£π¦ = 5 – velocity vector Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 ππ‘ π = π£π₯ π + π£π¦ π = −π + 5π π = π£π₯2 + π£π¦2 At π‘ = 1 π£π¦ = 5 – velocity vector Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 ππ‘ π = π£π₯ π + π£π¦ π = −π + 5π π = π£π₯2 + π£π¦2 = (−1)2 +(5)2 At π‘ = 1 π£π¦ = 5 – velocity vector Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 ππ‘ π = π£π₯ π + π£π¦ π = −π + 5π π = π£π₯2 + π£π¦2 = (−1)2 +(5)2 = 26 At π‘ = 1 π£π¦ = 5 – velocity vector Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 ππ‘ π = π£π₯ π + π£π¦ π = −π + 5π π = π£π₯2 + π£π¦2 = At π‘ = 1 π£π¦ = 5 – velocity vector (−1)2 +(5)2 = 26 – magnitude/speed Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 ππ‘ π = π£π₯ π + π£π¦ π = −π + 5π π = tan π = π£π₯2 + π£π¦2 = π£π¦ π£π₯ At π‘ = 1 π£π¦ = 5 – velocity vector (−1)2 +(5)2 = 26 – magnitude/speed Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ ππ‘ π = π£π₯ π + π£π¦ π = −π + 5π π = tan π = π£π₯2 + π£π¦2 = π£π¦ π£π₯ = 5 −1 At π‘ = 1 π£π¦ = 5 – velocity vector (−1)2 +(5)2 = 26 – magnitude/speed = −5 Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ ππ‘ – velocity vector π = π£π₯ π + π£π¦ π = −π + 5π π = tan π = π£π₯2 + π£π¦2 = π£π¦ π£π₯ = 5 −1 At π‘ = 1 π£π¦ = 5 (−1)2 +(5)2 = 26 – magnitude/speed = −5 π = tan−1 −5 Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ ππ‘ π = π£π₯ π + π£π¦ π = −π + 5π π = tan π = π£π₯2 + π£π¦2 = π£π¦ π£π₯ = 5 −1 At π‘ = 1 π£π¦ = 5 – velocity vector (−1)2 +(5)2 = 26 – magnitude/speed = −5 π = tan−1 −5 = −78°41′ Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ ππ‘ π = π£π₯ π + π£π¦ π = −π + 5π π = tan π = π£π₯2 + π£π¦2 = π£π¦ π£π₯ = 5 −1 – velocity vector (−1)2 +(5)2 = 26 – magnitude/speed = −5 π = 180° − 78°41′ At π‘ = 1 π£π¦ = 5 π = tan−1 −5 = −78°41′ Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ ππ‘ π = π£π₯ π + π£π¦ π = −π + 5π π = tan π = π£π₯2 + π£π¦2 = π£π¦ π£π₯ = 5 −1 At π‘ = 1 π£π¦ = 5 – velocity vector (−1)2 +(5)2 = 26 – magnitude/speed = −5 π = 180° − 78°41′ = 101°19′ π = tan−1 −5 = −78°41′ Exampl e: Find the magnitude and direction of velocity at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ ππ‘ π = π£π₯ π + π£π¦ π = −π + 5π π = tan π = π£π₯2 + π£π¦2 = π£π¦ π£π₯ = 5 −1 At π‘ = 1 π£π¦ = 5 – velocity vector (−1)2 +(5)2 = 26 – magnitude/speed = −5 π = 180° − 78°41′ = 101°19′ π = tan−1 −5 = −78°41′ – direction Acceleration in Curvilinear Motion Acceleration in Curvilinear Motion • Acceleration Vector π= ππ ππ‘ = π2 π ππ‘ 2 = π2 π₯ π ππ‘ 2 + π2 π¦ π ππ‘ 2 = ππ₯ π + ππ¦ π • Magnitude of of π π = πβπ= ππ₯2 + ππ¦2 • Direction ππ¦ tan ∅ = ππ₯ Acceleration in Curvilinear Motion • Acceleration Vector π= ππ ππ‘ = π2 π ππ‘ 2 = π2 π₯ π ππ‘ 2 + π2 π¦ π ππ‘ 2 = ππ₯ π + ππ¦ π • Magnitude of of π π = πβπ= ππ₯2 + ππ¦2 • Direction ππ¦ tan ∅ = ππ₯ Acceleration in Curvilinear Motion • Acceleration Vector π= ππ ππ‘ = π2 π ππ‘ 2 = π2 π₯ π ππ‘ 2 + π2 π¦ π ππ‘ 2 = ππ₯ π + ππ¦ π • Magnitude of of π π = πβπ= ππ₯2 + ππ¦2 • Direction ππ¦ tan ∅ = ππ₯ Exampl e: Exampl e: Find the magnitude and direction of acceleration at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. Exampl e: Find the magnitude and direction of acceleration at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π2π₯ ππ‘ 2 = ππ₯ = 0 π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 At π‘ = 1 = ππ¦ = 12π‘ ππ¦ = 12 ππ‘ π2π¦ ππ‘ 2 Exampl e: Find the magnitude and direction of acceleration at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π2π₯ ππ‘ 2 = ππ₯ = 0 π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 At π‘ = 1 = ππ¦ = 12π‘ ππ¦ = 12 ππ‘ π2π¦ ππ‘ 2 Exampl e: Find the magnitude and direction of acceleration at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π2π₯ ππ‘ 2 = ππ₯ = 0 π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 At π‘ = 1 = ππ¦ = 12π‘ ππ¦ = 12 ππ‘ π2π¦ ππ‘ 2 Exampl e: Find the magnitude and direction of acceleration at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π2π₯ ππ‘ 2 = ππ₯ = 0 π = ππ₯ π + ππ¦ π π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 At π‘ = 1 = ππ¦ = 12π‘ ππ¦ = 12 ππ‘ π2π¦ ππ‘ 2 Exampl e: Find the magnitude and direction of acceleration at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π2π₯ ππ‘ 2 = ππ₯ = 0 π = ππ₯ π + ππ¦ π = 12π π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 At π‘ = 1 = ππ¦ = 12π‘ ππ¦ = 12 ππ‘ π2π¦ ππ‘ 2 Exampl e: Find the magnitude and direction of acceleration at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π2π₯ ππ‘ 2 = ππ₯ = 0 π = ππ₯ π + ππ¦ π = 12π π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 At π‘ = 1 = ππ¦ = 12π‘ ππ¦ = 12 ππ‘ π2π¦ ππ‘ 2 – acceleration vector Exampl e: Find the magnitude and direction of acceleration at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π2π₯ ππ‘ 2 = ππ₯ = 0 π = ππ₯ π + ππ¦ π = 12π π = ππ₯2 + ππ¦2 π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 At π‘ = 1 = ππ¦ = 12π‘ ππ¦ = 12 ππ‘ π2π¦ ππ‘ 2 – acceleration vector Exampl e: Find the magnitude and direction of acceleration at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π2π₯ ππ‘ 2 π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 At π‘ = 1 = ππ¦ = 12π‘ ππ¦ = 12 ππ‘ π2π¦ ππ‘ 2 = ππ₯ = 0 π = ππ₯ π + ππ¦ π = 12π π = ππ₯2 + ππ¦2 = (0)2 +(12)2 = 12 – acceleration vector Exampl e: Find the magnitude and direction of acceleration at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π2π₯ ππ‘ 2 π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 At π‘ = 1 = ππ¦ = 12π‘ ππ¦ = 12 ππ‘ π2π¦ ππ‘ 2 = ππ₯ = 0 π = ππ₯ π + ππ¦ π = 12π π = ππ₯2 + ππ¦2 = (0)2 +(12)2 = 12 – acceleration vector – magnitude Exampl e: Find the magnitude and direction of acceleration at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π2π₯ ππ‘ 2 π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 At π‘ = 1 = ππ¦ = 12π‘ ππ¦ = 12 ππ‘ π2π¦ ππ‘ 2 = ππ₯ = 0 π = ππ₯ π + ππ¦ π = 12π π = tan ∅ = ππ₯2 + ππ¦2 = ππ¦ ππ₯ (0)2 +(12)2 = 12 – acceleration vector – magnitude Exampl e: Find the magnitude and direction of acceleration at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π2π₯ ππ‘ 2 π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 At π‘ = 1 = ππ¦ = 12π‘ ππ¦ = 12 ππ‘ π2π¦ ππ‘ 2 = ππ₯ = 0 – acceleration vector π = ππ₯ π + ππ¦ π = 12π π = tan ∅ = ππ₯2 + ππ¦2 = ππ¦ ππ₯ (0)2 +(12)2 = 12 – magnitude ∅ = 90° – direction Exampl e: Find the magnitude and direction of acceleration at time t, given π₯ = 2 − π‘, π¦ = 2π‘ 3 − π‘; at π‘ = 1. π₯ =2−π‘ ππ₯ = π£π₯ = −1 ππ‘ π2π₯ ππ‘ 2 π¦ = 2π‘ 3 − π‘ ππ¦ = π£π¦ = 6π‘ 2 − 1 At π‘ = 1 = ππ¦ = 12π‘ ππ¦ = 12 ππ‘ π2π¦ ππ‘ 2 = ππ₯ = 0 – acceleration vector π = ππ₯ π + ππ¦ π = 12π π = tan ∅ = ππ₯2 + ππ¦2 = ππ¦ ππ₯ (0)2 +(12)2 = 12 ∅ = 90° – magnitude Tangential and Normal Components of Acceleration Tangential and Normal Components of Acceleration • Tangent ππ‘ = π2 π ππ‘ 2 Tangential and Normal Components of Acceleration • Tangent ππ‘ = π2 π ππ‘ 2 • Normal ππ = π2 π or ππ2 = π 2 − ππ‘2 Exampl e: Exampl e: Find the magnitudes of the tangential and normal components of acceleration for the motion π₯ = π π‘ cos π‘, π¦ = π π‘ sin π‘ at any time π‘. Exampl e: Find the magnitudes of the tangential and normal components of acceleration for the motion π₯ = π π‘ cos π‘, π¦ = π π‘π₯sin time π‘. = ππ‘π‘ at cosany π‘ π¦ = π π‘ sin π‘ ππ₯ = π£π₯ = π π‘ cos π‘ − sin π‘ ππ‘ π2π₯ = ππ₯ = −2π π‘ sin π‘ 2 ππ‘ ππ¦ = π£π¦ = π π‘ sin π‘ + ππ‘ π2π¦ = ππ¦ = 2π π‘ cos π‘ 2 ππ‘ cos π‘ Exampl e: Find the magnitudes of the tangential and normal components of acceleration for the motion π₯ = π π‘ cos π‘, π¦ = π π‘π₯sin time π‘. = ππ‘π‘ at cosany π‘ π¦ = π π‘ sin π‘ ππ₯ = π£π₯ = π π‘ cos π‘ − sin π‘ ππ‘ π2π₯ = ππ₯ = −2π π‘ sin π‘ 2 ππ‘ ππ ππ‘ = π = π£π₯2 + π£π¦2 = ππ¦ = π£π¦ = π π‘ sin π‘ + ππ‘ π2π¦ = ππ¦ = 2π π‘ cos π‘ 2 ππ‘ ππ‘ cos π‘ − sin π‘ 2 cos π‘ + ππ‘ sin π‘ + cos π‘ 2 = 2π π‘ Exampl e: Find the magnitudes of the tangential and normal components of acceleration for the motion π₯ = π π‘ cos π‘, π¦ = π π‘π₯sin time π‘. = ππ‘π‘ at cosany π‘ π¦ = π π‘ sin π‘ ππ₯ = π£π₯ = π π‘ cos π‘ − sin π‘ ππ‘ π2π₯ = ππ₯ = −2π π‘ sin π‘ 2 ππ‘ ππ = π ππ‘ π2 π = ππ‘ 2 = 2π π‘ π£π₯2 + π£π¦2 = ππ¦ = π£π¦ = π π‘ sin π‘ + ππ‘ π2π¦ = ππ¦ = 2π π‘ cos π‘ 2 ππ‘ ππ‘ cos π‘ − sin π‘ 2 cos π‘ + ππ‘ sin π‘ + cos π‘ 2 = 2π π‘ Exampl e: Find the magnitudes of the tangential and normal components of acceleration for the motion π₯ = π π‘ cos π‘, π¦ = π π‘π₯sin time π‘. = ππ‘π‘ at cosany π‘ π¦ = π π‘ sin π‘ ππ₯ = π£π₯ = π π‘ cos π‘ − sin π‘ ππ‘ π2π₯ = ππ₯ = −2π π‘ sin π‘ 2 ππ‘ ππ = π ππ‘ π2 π = ππ‘ 2 ππ‘ = π£π₯2 + π£π¦2 = = ππ¦ = π£π¦ = π π‘ sin π‘ + ππ‘ π2π¦ = ππ¦ = 2π π‘ cos π‘ 2 ππ‘ ππ‘ cos π‘ − sin π‘ 2 cos π‘ + ππ‘ sin π‘ + cos π‘ 2 = 2π π‘ 2π π‘ π2π ππ‘ 2 = 2π π‘ – magnitudes of the tangential Exampl e: Find the magnitudes of the tangential and normal components of acceleration for the motion π₯ = π π‘ cos π‘, π¦ = π π‘π₯sin time π‘. = ππ‘π‘ at cosany π‘ π¦ = π π‘ sin π‘ ππ₯ = π£π₯ = π π‘ cos π‘ − sin π‘ ππ‘ π2π₯ = ππ₯ = −2π π‘ sin π‘ 2 ππ‘ π2π ππ‘ = 2 = 2π π‘ ππ‘ ππ¦ = π£π¦ = π π‘ sin π‘ + ππ‘ π2π¦ = ππ¦ = 2π π‘ cos π‘ 2 ππ‘ cos π‘ Exampl e: Find the magnitudes of the tangential and normal components of acceleration for the motion π₯ = π π‘ cos π‘, π¦ = π π‘π₯sin time π‘. = ππ‘π‘ at cosany π‘ π¦ = π π‘ sin π‘ ππ₯ = π£π₯ = π π‘ cos π‘ − sin π‘ ππ‘ π2π₯ = ππ₯ = −2π π‘ sin π‘ 2 ππ‘ π2π ππ‘ = 2 = 2π π‘ ππ‘ π = ππ₯2 + ππ¦2 = −2π π‘ sin π‘ ππ¦ = π£π¦ = π π‘ sin π‘ + ππ‘ π2π¦ = ππ¦ = 2π π‘ cos π‘ 2 ππ‘ 2 + 2π π‘ cos π‘ 2 cos π‘ = 2π π‘ Exampl e: Find the magnitudes of the tangential and normal components of acceleration for the motion π₯ = π π‘ cos π‘, π¦ = π π‘π₯sin time π‘. = ππ‘π‘ at cosany π‘ π¦ = π π‘ sin π‘ ππ₯ = π£π₯ = π π‘ cos π‘ − sin π‘ ππ‘ π2π₯ = ππ₯ = −2π π‘ sin π‘ 2 ππ‘ π2π ππ‘ = 2 = 2π π‘ ππ‘ π = ππ2 ππ₯2 + ππ¦2 = 2 = π − ππ‘2 ππ¦ = π£π¦ = π π‘ sin π‘ + ππ‘ π2π¦ = ππ¦ = 2π π‘ cos π‘ 2 ππ‘ −2π π‘ sin π‘ 2 + 2π π‘ cos π‘ ππ = 2 π − 2 ππ‘2 cos π‘ = 2π π‘ = 2π π‘ 2 − 2π π‘ 2 Exampl e: Find the magnitudes of the tangential and normal components of acceleration for the motion π₯ = π π‘ cos π‘, π¦ = π π‘π₯sin time π‘. = ππ‘π‘ at cosany π‘ π¦ = π π‘ sin π‘ ππ₯ = π£π₯ = π π‘ cos π‘ − sin π‘ ππ‘ π2π₯ = ππ₯ = −2π π‘ sin π‘ 2 ππ‘ π2π ππ‘ = 2 = 2π π‘ ππ‘ π = ππ2 ππ₯2 + ππ¦2 = 2 = π − ππ‘2 ππ = 2π π‘ ππ¦ = π£π¦ = π π‘ sin π‘ + ππ‘ π2π¦ = ππ¦ = 2π π‘ cos π‘ 2 ππ‘ −2π π‘ sin π‘ 2 + 2π π‘ cos π‘ ππ = 2 π − 2 ππ‘2 cos π‘ = 2π π‘ = 2π π‘ 2 − – magnitudes of the normal 2π π‘ 2 Assessmen t: 1. 2. Find the magnitude and direction of velocity and acceleration of time π‘ , given π₯ = π cos 3π‘, π¦ = sin π‘ at π‘ = . 4 Find the magnitude and direction of velocity and acceleration for the motion π₯ = 3π‘ and π¦ = 9π‘ − 3π‘ 2 at π‘ = 2.