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Distance, Speed and Acceleration

Learn how rates of change apply to displacement, velocity, and time in motion equations to calculate displacement, velocity, and acceleration at specific time intervals and analyze objects' motion behaviors.

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Distance, Speed and Acceleration

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  1. Distance, Speed andAcceleration Special Rates of Change

  2. Displacement, Velocity and Acceleration Special Rates of Change

  3. What is to be learned? • How rates of change apply to displacement, velocity and time.

  4. x Displacement Distance (in a certain direction) Velocity Change in displacement over a period of time Acceleration Change in velocity over a period of time (some call it speed!) v = dx/dt a = dv/dt

  5. Ex An object is moving along x axis (cm) at time t(secs) according to equation: x = 4t2 + 10t -t3 Calculate: a) Displacement, Velocity and Acceleration after 1 second and 3 seconds b) Displacement and Velocity after 5 seconds

  6. x = 4t2 + 10t -t3 displacement t = 1 x = 4(1)2 + 10(1) – 13 = 13cm velocity v = dx/dt = 8t + 10 – 3t2 t = 1 v = 8(1) + 10 – 3(1)2 = 15 cm/sec

  7. x = 4t2 + 10t -t3 velocity v = dx/dt = 8t + 10 – 3t2 t = 1 v = 8(1) + 10 – 3(1)2 = 15 cm/sec acceleration a = dv/dt = 8 – 6t t = 1 a = 8 – 6(1) = 2 cm/sec/sec cm/sec2

  8. x = 4t2 + 10t -t3 displacement t = 3 x = 4(3)2 + 10(3) – 33 = 39cm velocity v = dx/dt = 8t + 10 – 3t2 t = 3 v = 8(3) + 10 – 3(3)2 = 7 cm/sec

  9. x = 4t2 + 10t -t3 velocity v = dx/dt = 8t + 10 – 3t2 t = 3 v = 8(3) + 10 – 3(3)2 = 7 cm/sec acceleration a = dv/dt = 8 – 6t t = 3 a = 8 – 6(3) = -10 cm/sec2

  10. x = 4t2 + 10t -t3 velocity v = dx/dt = 8t + 10 – 3t2 t = 3 v = 8(3) + 10 – 3(3)2 = 7 cm/sec acceleration a = dv/dt = 8 – 6t t = 3 a = 8 – 6(3) = -10 cm/sec2 It is decelerating

  11. x = 4t2 + 10t -t3 displacement t = 5 x = 4(5)2 + 10(5) – 53 = 25cm velocity v = dx/dt = 8t + 10 – 3t2 t = 5 v = 8(5) + 10 – 3(5)2 = -25 cm/sec

  12. x = 4t2 + 10t -t3 displacement t = 5 x = 4(5)2 + 10(5) – 53 = 25cm velocity v = dx/dt = 8t + 10 – 3t2 t = 5 v = 8(5) + 10 – 3(5)2 = -25 cm/sec It has changed direction

  13. Motion and Derivatives x (or h) Displacement Distance (in a certain direction) Velocity Change in displacement over a period of time Acceleration Change in velocity over a period of time (some call it speed!) v = dx/dt (0r dh/dt) a = dv/dt

  14. Ex Belinda throws a ball into the air Its height (h m) after t secs is: h = 4t – t2 Find a) Its height, velocity and acceleration after 1 sec b) What is its height when the velocity is zero?

  15. h = 4t – t2 height (displacement) t = 1 h = 4(1) – 12 = 3m velocity v = dh/dt = 4 – 2t t = 1 v = 4 – 2(1) = 2m/sec

  16. h = 4t – t2 velocity v = dh/dt = 4 – 2t t = 1 v = 4 – 2(1) = 2m/sec acceleration a = dv/dt = -2 t = 1 (or whatever!) a = -2 m/sec2 decelerating by 2m/sec2

  17. b) Velocity zero? v = dh/dt = 4 – 2t  4 – 2t = 0 4 = 2t t = 2 velocity is 0 m/sec after 2 seconds height? h = 4t – t2 t = 2 h = 4(2) – 22 = 4 m

  18. Graphical explanation v = 0 h = 4t – t2 h 4 3 1 2 t

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