Distance Speed and Acceleration Special Rates of Change

Distance, Speed and Acceleration Special Rates of Change

Displacement, Velocity and Acceleration Special Rates of Change

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

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

Ex An object is moving along x axis (cm) at time t(secs) according to equation: x = 4 t 2 + 10 t - t 3 Calculate: a) Displacement, Velocity and Acceleration after 1 second and 3 seconds b) Displacement and Velocity after 5 seconds

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

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

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

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

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

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

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

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

Ex Belinda throws a ball into the air Its height (h m) after t secs is: h = 4 t – t 2 Find a) Its height, velocity and acceleration after 1 sec b) What is its height when the velocity is zero?

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

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

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

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