Equations of motion • Displacement, velocity & acceleration • Velocity-time graph • The five kinematics equations • Solving kinematics problems • Gravity Equations of motion - 1 VCE Physics.com Displacement, velocity & acceleration • Displacement is a vector quantity - it is the change of position of an object. • Velocity is a vector quantity - it is the rate of change of displacement with time. • Acceleration is the rate of change of velocity with time. • There are five kinematics rules: each one requires three of the variables to be known. • These rules are based on constant acceleration. • These rules apply to motion in one dimension (forward & back, up & down....). Directions can defined as positive or negative. x = displacement (m) t = time (s) a = acceleration (m/s2) u = initial velocity (m/s) v = final velocity (m/s) Equations of motion - 2 VCE Physics.com Velocity-time graphs • The gradient of the graph gives the acceleration. • The area under the graph gives the displacement. Velocity (m/s) v Acceleration u Time (s) t Displacement 1 m/s x 1 s = 1m Equations of motion - 3 VCE Physics.com Defining the equations (1) • The gradient of the graph gives the acceleration. Velocity (m/s) v v - u v-u a = t u t Time (s) t Equations of motion - 4 VCE Physics.com Defining the equations (2) • The area under the graph gives the displacement. Velocity (m/s) Trapezium area v 1 = (u +v)t 2 u 1 v x = (u +v)t 2 u t Time (s) Equations of motion - 5 VCE Physics.com Defining the equations (3) • The area under the graph gives the displacement. Triangle area 1 = t × at Velocity (m/s) 2 1 = at2 v 1 2 at 2 2 v - u = at u 1 x = ut+ at2 2 ut Time (s) t Equations of motion - 6 VCE Physics.com Defining the equations (4) • The area under the graph gives the displacement. 1 2 Velocity (m/s) at 2 v 1 x = vt- at2 2 u vt Time (s) t Equations of motion - 7 VCE Physics.com Defining the equations (5) • We can also define a rule that is not dependent on time. v-u v-u a = t = t a 1 (u +v) (v −u) x = (u +v)t x = 2 2 a v 2 −u2 x = 2a Equations of motion - 8 VCE Physics.com Solving kinematics problems (1) • A car accelerates from the lights, 0 to 60 km/h in 4.0 seconds • Find the distance covered in that time. Known information: 1 x = (u +v)t 2 u = 0 m/s 60 km/h 1 v = = 17 m/s x = (0 + 17 m/s) × 4 s 3.6 2 t = 4.0 s x = 34 m (This is an average speed of 8.5 m/s over the 4 seconds.) Equations of motion - 9 VCE Physics.com Solving kinematics problems (2) • The brakes & tyres of a car can provide a maximum deceleration of around 6 m/s2. If a car is traveling at 100 km/h, what is the minimum distance in which it could stop? Known information: v2 - u2 100 km/h x = u = = 28 m/s 2a 3.6 (0 m/s)2 - (28 m/s)2 2 x = a = -6 m/s 2 2 × -6 m/s v = 0 m/s x =65 m (Stopping distance varies with the square of speed. 2 x speed = 4 x stopping distance.) Equations of motion - 10 VCE Physics.com Solving kinematics problems (3) • A tennis ball is hit upwards at a speed of 15 m/s. • How high does the ball go? How long does it take to hit the ground? v-u v-u Known information: a = t = t a u = 15 m/s - 15 m/s - 15 m/s t = 2 2 -10 m/s a = -10 m/s t = 3 s Acceleration due to gravity Maximum height at 1.5 s, v = 0 2 g ≈10 m/s 1 x = (u +v)t If the ball lands at the height 2 1 that it was hit from: x = (15 m/s + 0 m/s)(1.5 s) v = -15 m/s 2 x = 11.25 m Equations of motion - 11 VCE Physics.com Gravity • Objects falling (or rising up) experience an 0 m acceleration due to the force of gravity. 5 m • g (on Earth) = 9.81 m/s2 ≈10 m/s2 20 m • (Presuming that air resistance is insignificant: the projectile is heavy, small & not moving very fast.) 45 m Time (s) Speed Distance 0 0 m/s 0 m u = 0 m /s 80 m 1 10 m/s 5 m v = at 2 20 m/s 20 m 1 x = at2 3 30 m/s 45 m 2 4 40 m/s 80 m 125 m 5 50 m/s 125 m Equations of motion - 12 VCE Physics.com Graphs of motion under gravity Acceleration a = constant time (s) u = 0 m /s v = u + at a = -10 m/s2 1 x =ut + at2 2 Velocity time (s) v = −10t x = −5t2 Equations of motion - 13 VCE Physics.com.