MOTION MOTION © John Parkinson

Velocity = Speed in a Specified Direction Constant Velocity Distance travelled - s Time taken - t Velocity - v s v = v t s / t Velocity = Speed in a Specified Direction © John Parkinson

PHYSICS BUS N 100 m in 4 seconds Distance travelled = ? 100 m Displacement = ? 100 m to the East Speed = ? Speed = 100/4 = 25 m s-1 Velocity = 25 m s-1 to the East Velocity = ? © John Parkinson

DISPLACEMENT – TIME GRAPHS Constant velocity Displacement - s What will the graph look like? Δs GRADIENT = ? Δt Time - t VELOCITY © John Parkinson

What about this graph? And this graph? Displacement - s Time - t What about this graph? A body at rest Displacement - s Time - t Δs Δt And this graph? The gradient is …….? increasing The body must be ……..? accelerating © John Parkinson

This body has a constant or uniform ………? VELOCITY – TIME GRAPHS Velocity - v Time - t This body has a constant or uniform ………? Δv Δt acceleration The gradient = ? the acceleration 1 = …… ? Uniform acceleration Velocity - v Time - t 2 = …… ? Constant velocity 2 A 1 3 = …… ? Uniform retardation [deceleration] 3 Area under the graph = A = …….. ? DISTANCE TRAVELLED © John Parkinson

QUESTION The graph represents the motion of a tube train between two stations Find The acceleration The maximum velocity The retardation The distance travelled Velocity – v/ms-1 Time – t/s 30 20 50 80 1. The acceleration = the initial gradient = 30÷20 = 1.5 m s-2 2. The maximum velocity is read from the graph = 30 m s-1 3. The retardation = the final gradient = -30 ÷ [80-50] = -1.0 m s-2 The distance travelled = the area under the graph =½ x 20 x 30 + 30 x 30 + ½ x 30 x 30 = 1650 m © John Parkinson

s1 What will the distance – time, velocity - time and acceleration time graphs look like for this bouncing ball? Displacement - s Time - t s1 s2 s2 Velocity - v Time - t © John Parkinson

9.81ms-2 © John Parkinson Velocity - v Time - t Acceleration - a

What might this graph represent? Velocity What might this graph represent? Terminal Velocity Time Can you draw an acceleration time graph for this motion? Acceleration Time 9.81 m s-1 © John Parkinson

EQUATIONS OF UNIFORM ACCELERATION DISTANCE = VELOCITY x TIME EQUATIONS OF MOTION EQUATIONS OF UNIFORM ACCELERATION For Constant Velocity VELOCITY = DISTANCE TIME If Velocity is not constant , this equation just gives the average velocity for the journey DISTANCE = VELOCITY x TIME © John Parkinson

For UNIFORM ACCELERATION EQUATIONS OF MOTION EQUATIONS OF MOTION For UNIFORM ACCELERATION SYMBOLS a = ACCELERATION u = INITIAL VELOCITY v = FINAL VELOCITY s = DISTANCE TRAVELLED t = TIME TAKEN © John Parkinson

For UNIFORM ACCELERATION 1. Distance travelled = average velocity times the time taken s = u + v 2 t Velocity Time u v t 2. Acceleration = the change in velocity per second a = v - u t Rearranging v = u + at © John Parkinson

3. Substituting equation [2] into equation [1] u + v 2 t v = u + at s = ut + at2 1 2 4. Rearrange equation 1. to make t the subject Now substitute this in equation 3 and rearrange to give : v2 = u2 + 2as © John Parkinson

USING THE EQUATIONS OF MOTION 1. Write down the symbols of the quantities that you know 2. Write down the symbol of the quantity that you require 3. Select the equation that contains all of the symbols in 1. and 2. above e.g. A stone is released from a height of 20 m above the ground. Neglecting air resistance and using the acceleration due to gravity as 9.81 ms-2, find the velocity with which the stone will hit the ground . This must be equation 2 as it is the only one with “v”, “u”, “a” and “s” in it u = 0 from rest s = 20 m v2 = u2 + 2as a = 9.81 ms-2 v2 = 02 + 2 x 9.81 x 20 v = ? v = 392 = 19.8 m s-1 © John Parkinson