Distance, Speed and Acceleration

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