Physics 13 General Physics 1 Kinematics of Motion Motion in Two or Three Dimensions MARLON FLORES SACEDON

What is Instantaneous velocity? x𝑡-Graph Motion in Two or Three Dimensions: Instantaneous Velocity and Acceleration What is Instantaneous velocity? x𝑡-Graph Time (𝑡) Position (𝑥) 𝑥 𝑓 𝑡 𝑓 ∆𝑥 = 𝑥 𝑓 − 𝑥 𝑖 𝑥 𝑖 𝑡 𝑖 ∆𝑡 = 𝑡 𝑓 − 𝑡 𝑖 ∆𝑥 ∆𝑡 = 𝑥 𝑓 − 𝑥 𝑖 𝑡 𝑓 − 𝑡 𝑖 𝒗 𝒂𝒗 = Average velocity

What is Instantaneous velocity? x𝑡-Graph Motion in Two or Three Dimensions: Instantaneous Velocity and Acceleration What is Instantaneous velocity? x𝑡-Graph Time (𝑡) Position (𝑥) 𝑥 𝑓 𝑥 𝑓 Slope of the secant line ∆𝑥 𝑡 𝑓 ∆𝑥 𝑡 𝑓 Slope of the secant line Slope of the secant line 𝑥 𝑓 ∆𝑥 𝑡 𝑓 𝑥 𝑖 𝑡 𝑖 Slope of the tangent line ∆𝑥 ∆𝑡 ∆𝑡 ∆𝑥 ∆𝑡 = 𝑥 𝑓 − 𝑥 𝑖 𝑡 𝑓 − 𝑡 𝑖 𝒗 𝒂𝒗 = Average velocity = lim ∆𝒕→𝟎 ∆𝑥 ∆𝑡 = 𝒅𝒙 𝒅𝒕 𝒗 𝒊𝒏𝒔 = lim ∆𝒕→𝟎 𝒗 𝒂𝒗 Instantaneous velocity = lim ∆𝒕→𝟎 ∆v ∆𝑡 = 𝒅𝒗 𝒅𝒕 𝒂 𝒊𝒏𝒔 = lim ∆𝒕→𝟎 𝒂 𝒂𝒗 Instantaneous acceleration

What is Instantaneous velocity? x𝑡-Graph v𝑡-Graph Motion in Two or Three Dimensions: Instantaneous Velocity and Acceleration What is Instantaneous velocity? x𝑡-Graph Time (𝑡) Velocity (𝑣) v𝑡-Graph Time (𝑡) Position (𝑥) Average acceleration 𝑎 𝑖𝑛𝑠 𝑣 𝑖𝑛𝑠 Average velocity 𝑣 𝑖𝑛𝑠 = 𝑑𝑥 𝑑𝑡 𝑣 𝑖𝑛𝑠 = 𝑑𝑣 𝑑𝑡 𝑎 𝑖𝑛𝑠 ∆𝑥 ∆𝑡 = 𝑥 𝑓 − 𝑥 𝑖 𝑡 𝑓 − 𝑡 𝑖 𝒗 𝒂𝒗 = Average velocity = lim ∆𝒕→𝟎 ∆𝑥 ∆𝑡 = 𝒅𝒙 𝒅𝒕 𝒗 𝒊𝒏𝒔 = lim ∆𝒕→𝟎 𝒗 𝒂𝒗 Instantaneous velocity = lim ∆𝒕→𝟎 ∆v ∆𝑡 = 𝒅𝒗 𝒅𝒕 𝒂 𝒊𝒏𝒔 = lim ∆𝒕→𝟎 𝒂 𝒂𝒗 Instantaneous acceleration

Position & Velocity Vectors Motion in Two or Three Dimensions: The Position, Velocity, and Acceleration Position & Velocity Vectors 𝑧 𝑥 𝑦 𝑟 =𝑥 𝑖 +𝑦 𝑗 +𝑧 𝑘 𝑧 𝑘 𝑦 𝑗 Where: 𝑟 is a position vector x, y, & z are magnitudes of the components of position vector r 𝑖 , 𝑗 , & 𝑘 are unit vectors 𝑝 𝑟 𝑥 𝑖

Motion in Two or Three Dimensions: The Position, Velocity, and Acceleration 𝑧 𝑥 𝑦 ∆ 𝑟 = 𝑥 2 − 𝑥 1 𝑖 + 𝑦 2 − 𝑦 1 𝑗 + 𝑧 2 − 𝑧 1 𝑘 Where: ∆ 𝑟 is the change in position vector 𝑟 2 𝑝 2 𝑣 𝑎𝑣𝑒 = ∆ 𝑟 ∆𝑡 = 𝑟 2 − 𝑟 1 𝑡 2 − 𝑡 1 ∆ 𝑟 = lim ∆𝑡→0 𝑟 2 − 𝑟 1 𝑡 2 − 𝑡 1 𝑣 = lim ∆𝑡→0 𝑣 𝑎𝑣𝑒 𝑟 1 𝑝 1 𝑣 = lim ∆𝑡→0 ∆ 𝑟 ∆𝑡 = 𝑑 𝑟 𝑑𝑡 But: 𝑣 = 𝑣 𝑥 𝑖 + 𝑣 𝑦 𝑗 + 𝑣 𝑧 𝑘 𝑣 𝑥 = 𝑑𝑥 𝑑𝑡 𝑣 𝑦 = 𝑑𝑦 𝑑𝑡 𝑣 𝑧 = 𝑑𝑧 𝑑𝑡 𝑟 1 = 𝑥 1 𝑖 + 𝑦 1 𝑗 + 𝑧 1 𝑘 𝑟 2 = 𝑥 2 𝑖 + 𝑦 2 𝑗 + 𝑧 2 𝑘 𝑣 = 𝑑𝑥 𝑑𝑡 𝑖 + 𝑑𝑦 𝑑𝑡 𝑗 + 𝑑𝑧 𝑑𝑡 𝑘 ∆ 𝑟 = 𝑟 2 − 𝑟 1 ∆ 𝑟 = 𝑥 2 𝑖 + 𝑦 2 𝑗 + 𝑧 2 𝑘 − 𝑥 1 𝑖 + 𝑦 1 𝑗 + 𝑧 1 𝑘 𝑣= 𝑣 𝑥 2 + 𝑣 𝑦 2 + 𝑣 𝑧 2

Motion in Two or Three Dimensions: The Position, Velocity, and Acceleration 𝑎 = 𝑑 𝑣 𝑑𝑡 = 𝑑 𝑣 𝑥 𝑑𝑡 𝑖 + 𝑑 𝑣 𝑦 𝑑𝑡 𝑗 + 𝑑 𝑣 𝑧 𝑑𝑡 𝑘 𝑎 = 𝑑 2 𝑟 𝑑𝑡 2 = 𝑑 2 𝑥 𝑑𝑡 2 𝑖 + 𝑑 2 𝑦 𝑑𝑡 2 𝑗 + 𝑑 2 𝑧 𝑑𝑡 2 𝑘

Motion in Two or Three Dimensions: The Position, Velocity, and Acceleration Problem # 1

Motion in Two or Three Dimensions: The Position, Velocity, and Acceleration PROBLEM: A robotic vehicle, or rover, is exploring the surface of Mars. The stationary Mars lander is the origin of coordinates, and the surrounding Martian surface lies in the xy-plane. The rover, which we represent as a point, has x- and y-coordinates that vary with time: 𝑥=2.0𝑚− 0.25 𝑚 𝑠 2 𝑡 2 𝑦= 1.0 𝑚 𝑠 𝑡+ 0.025 𝑚 𝑠 2 𝑡 3 (a) Find the rover’s coordinates and distance from the lander at t = 2.0 s. (b) Find the rover’s displacement and average velocity vectors for the interval t = 0.0 s to t = 2.0 s. (c) Find a general expression for the rover’s instantaneous velocity vector 𝑣 . Express 𝑣 at t = 2.0 s in component form and in terms of magnitude and direction. (a) solution @ 𝑡=2.0𝑠 (b) solution ∆ 𝑟 = 𝑟 2 − 𝑟 1 @ 𝑡=0𝑠 = 1.0𝑚 𝑖 + 2.2𝑚 𝑗 − 2.0𝑚 𝑖 𝑥=2.0𝑚− 0.25 𝑚 𝑠 2 𝑡 2 𝑥=2.0𝑚− 0.25 𝑚 𝑠 2 𝑡 2 =− 1.0𝑚 𝑖 + 2.2𝑚 𝑗 =2.0𝑚− 0.25 𝑚 𝑠 2 2𝑠 2 =2.0𝑚− 0.25 𝑚 𝑠 2 0𝑠 2 𝑣 𝑎𝑣𝑒 = ∆ 𝑟 ∆𝑡 = − 1.0𝑚 𝑖 + 2.2𝑚 𝑗 2.0𝑠−0𝑠 =1.0𝑚 =2.0𝑚 𝑦= 1.0 𝑚 𝑠 𝑡+ 0.025 𝑚 𝑠 2 𝑡 3 𝑦= 1.0 𝑚 𝑠 𝑡+ 0.025 𝑚 𝑠 2 𝑡 3 =− 0.5𝑚 𝑖 + 1.1𝑚 𝑗 = 1.0 𝑚 𝑠 2𝑠 + 0.025 𝑚 𝑠 2 2𝑠 3 = 1.0 𝑚 𝑠 0𝑠 + 0.025 𝑚 𝑠 2 0𝑠 3 = 1.0 𝑚 𝑠 2𝑠 + 0.025 𝑚 𝑠 2 2𝑠 3 =0 =2.2𝑚 @ 𝑡=0𝑠: 𝑟 1 = 2.0𝑚 𝑖 𝑟= 𝑥 2 + 𝑦 2 = 𝑥 1.0𝑚 2 + 2.2𝑚 2 =2.4𝑚 @ 𝑡=2.0𝑠: 𝑟 2 = 1.0𝑚 𝑖 + 2.2𝑚 𝑗

Motion in Two or Three Dimensions: The Position, Velocity, and Acceleration @ 𝑡=2 𝑠 PROBLEM: A robotic vehicle, or rover, is exploring the surface of Mars. The stationary Mars lander is the origin of coordinates, and the surrounding Martian surface lies in the xy-plane. The rover, which we represent as a point, has x- and y-coordinates that vary with time: 𝒓 𝟐 ∆ 𝒓 𝑥=2.0𝑚− 0.25 𝑚 𝑠 2 𝑡 2 𝑦= 1.0 𝑚 𝑠 𝑡+ 0.025 𝑚 𝑠 2 𝑡 3 (a) Find the rover’s coordinates and distance from the lander at t = 2.0 s. (b) Find the rover’s displacement and average velocity vectors for the interval t = 0.0 s to t = 2.0 s. (c) Find a general expression for the rover’s instantaneous velocity vector 𝑣 . Express 𝑣 at t = 2.0 s in component form and in terms of magnitude and direction. 𝒓 𝟏 @ 𝑡=0 (a) solution @ 𝑡=2.0𝑠 (b) solution ∆ 𝑟 = 𝑟 2 − 𝑟 1 @ 𝑡=0𝑠 = 1.0𝑚 𝑖 + 2.2𝑚 𝑗 − 2.0𝑚 𝑖 𝑥=2.0𝑚− 0.25 𝑚 𝑠 2 𝑡 2 𝑥=2.0𝑚− 0.25 𝑚 𝑠 2 𝑡 2 =− 1.0𝑚 𝑖 + 2.2𝑚 𝑗 =2.0𝑚− 0.25 𝑚 𝑠 2 2𝑠 2 =2.0𝑚− 0.25 𝑚 𝑠 2 0𝑠 2 𝑣 𝑎𝑣𝑒 = ∆ 𝑟 ∆𝑡 = − 1.0𝑚 𝑖 + 2.2𝑚 𝑗 2.0𝑠−0𝑠 =1.0𝑚 =2.0𝑚 𝑦= 1.0 𝑚 𝑠 𝑡+ 0.025 𝑚 𝑠 2 𝑡 3 𝑦= 1.0 𝑚 𝑠 𝑡+ 0.025 𝑚 𝑠 2 𝑡 3 =− 0.5𝑚 𝑖 + 1.1𝑚 𝑗 = 1.0 𝑚 𝑠 2𝑠 + 0.025 𝑚 𝑠 2 2𝑠 3 = 1.0 𝑚 𝑠 0𝑠 + 0.025 𝑚 𝑠 2 0𝑠 3 = 1.0 𝑚 𝑠 2𝑠 + 0.025 𝑚 𝑠 2 2𝑠 3 =0 =2.2𝑚 @ 𝑡=0𝑠: 𝑟 1 = 2.0𝑚 𝑖 𝑟= 𝑥 2 + 𝑦 2 = 𝑥 1.0𝑚 2 + 2.2𝑚 2 =2.4𝑚 @ 𝑡=2.0𝑠: 𝑟 2 = 1.0𝑚 𝑖 + 2.2𝑚 𝑗

Motion in Two or Three Dimensions: The Position, Velocity, and Acceleration 𝑥=2.0𝑚− 0.25 𝑚 𝑠 2 𝑡 2 @ 𝑡=2.0𝑠 @ 𝑡=0 @ 𝑡=2 𝑠 𝒓 𝟐 𝒓 𝟏 ∆ 𝒓 𝑣=1.64𝑚/𝑠 𝑣 𝑦 52.4 𝑜 −𝑣 𝑥 PROBLEM: A robotic vehicle, or rover, is exploring the surface of Mars. The stationary Mars lander is the origin of coordinates, and the surrounding Martian surface lies in the xy-plane. The rover, which we represent as a point, has x- and y-coordinates that vary with time: 𝑥=2.0𝑚− 0.25 𝑚 𝑠 2 𝑡 2 𝑦= 1.0 𝑚 𝑠 𝑡+ 0.025 𝑚 𝑠 2 𝑡 3 (a) Find the rover’s coordinates and distance from the lander at t = 2.0 s. (b) Find the rover’s displacement and average velocity vectors for the interval t = 0.0 s to t = 2.0 s. (c) Find a general expression for the rover’s instantaneous velocity vector 𝑣 . Express 𝑣 at t = 2.0 s in component form and in terms of magnitude and direction. 𝑣 = −0.50 𝑚 𝑠 2 𝑡 𝑖 + 1 𝑚 𝑠 + 0.075 𝑚 𝑠 2 𝑡 2 𝑗 (c) solution 𝑣 = 𝑣 𝑥 𝑖 + 𝑣 𝑦 𝑗 + 𝑣 𝑧 𝑘 𝑣 𝑥 = −0.50 𝑚 𝑠 2 𝑡 𝑣 = 𝑑𝑥 𝑑𝑡 𝑖 + 𝑑𝑦 𝑑𝑡 𝑗 + 𝑑𝑧 𝑑𝑡 𝑘 𝑣 𝑦 =1 𝑚 𝑠 + 0.075 𝑚 𝑠 2 𝑡 2 = 𝑑 2.0𝑚− 0.25 𝑚 𝑠 2 𝑡 2 𝑑𝑡 𝑣 𝑥 = 𝑑𝑥 𝑑𝑡 = −0.50 𝑚 𝑠 2 𝑡 @ t=2s: 𝑣 𝑥 = −0.50 𝑚 𝑠 2 2=−1.0𝑚/𝑠 𝑣 𝑦 =1 𝑚 𝑠 + 0.075 𝑚 𝑠 2 2 2 =1.3𝑚/𝑠 = 𝑑 1.0 𝑚 𝑠 𝑡+ 0.025 𝑚 𝑠 2 𝑡 3 𝑑𝑡 𝑣 𝑦 = 𝑑𝑦 𝑑𝑡 =1 𝑚 𝑠 + 0.075 𝑚 𝑠 2 𝑡 2 𝑣= 𝑣 𝑥 2 + 𝑣 𝑥 2 = −1.0𝑚/𝑠 2 + 1.3𝑚/𝑠 2 =1.64𝑚/𝑠 = 𝑇𝑎𝑛 −1 1.3 1.0 𝛾= 𝑇𝑎𝑛 −1 𝑣 𝑦 𝑣 𝑥 = 52.4 𝑜 𝑣 𝑧 = 𝑑𝑧 𝑑𝑡 = 𝑑 0 𝑑𝑡 =0

Motion in Two or Three Dimensions: The Position, Velocity, and Acceleration

Motion in Two or Three Dimensions: The Position, Velocity, and Acceleration

Motion in Two or Three Dimensions: The Position, Velocity, and Acceleration

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