On the move

• Distance- The space between two fixed positions
• Displacement- The distance between two fixed points and the direction
• Speed- The rate of change of distance
• Velocity- The rate of change of displacement

• V = S/T
• For any circle you can find the distance around the circumference using v = (2πr)/t
• An object undergoing circular motion is at a constant speed but it's direction is changing. This means that it's velocity is changing.
• A change in velocity requires a force to act on the body which in turn causes an acceleration. This must be towards the centre of the circle.

Instantaneous Speed

• If you want to know the speed of an object at a certain time, when the object is changing speed then you need to use the same formula but over an infinitesimal time period v = change in S / change in T
• This can be done in the real world by taking two pictures very close together in time and measuring the small distance change in the object

Finding instantaneous speed from graph

• Draw tangent onto the graph
• Calculate gradient of the tangent

• Accelerartion is the rate of change of velocity
• Acceleration = (Change in velocity) / time
• Deceleration is considered as negative acceleration in physics

Rate of Change

• Recall that the rate of change of a graph is the gradient of the graph
• By plotting a velocity-time graph you can calculate the acceleration by taking the gradient

Finding Instantaneous acceleration from a curved graph

• If the acceleration is changing over time then velocity-time graph will not be linear
• The gradient is calculated by taking the tangent of the curve at the time in question

• Consider an object that is accelerating from initial speed u to final velocity in time t. You can calculate the rate of change of velocity given by: A = (V-U)/T and V = U + AT
• The displacement S is therefore calculated by average speed multiplied by time T:                   S = ((V-U)T)/2

Combining These Formulae together you create:

• S = ((U+U+AT)T)/2 = S= (2UT+AT^2)/2 = S= UT + (AT^2)/2

Eliminating Time

• AS = (V-U)/T x ((U+V)T)/2
• AS = ((V-U)(V+U))/2 = (V^2 - U^2)/2
• V^2 = U^2 + 2AS

Area under a speed-time graph

• The area under a speed-time graph is the total distance covered S
• These should always be triangles, rectangles or trapeziums. In the case of curves, an estimate will need to be taken by counting the squares

• Free Fall- Constant acceleration without resistive force
• All objects on earth fall at a constant acceleration (g) equal to 9.81ms^-1
• S = UT + 1/2AT^2              V^2 = U^2 + 2AS
• In both of these acceleration formulae there is no mention of mass, in other words all objects fall at the same rate irrespective of mass so long as no other forces act

• Many problems start with an object undergoing uniform acceleration and then having that acceleration change in magnitude and direction at a future time

The key to answering these questions is two realise that:

• It is in two seperate parts. Work the first part out without considering the second section
• The final velocity of section one will pass on to become the initial velocity of the second section

• A projectile is an unpowered moving object within Earth gravity with negligable air resistance
• Only gravity acts as a force on the object so the only acceleration is downwards
• Acceleration in the vertical componenet is always g which is 9.81ms^-1
• The horizontal velocty of the object is a constant
• The motion in the vertical and the horizontal are completely independent of each other

Vertical Projection

• When an object is thrown upwards vertically then it will slow down, stop and then continue to accelerate to the floor
• The velocity at any time can be calculated using V = U - GT

• The height at any point is Sy = UT - 1/2GT^2
• Sy is the vertical distance and Sx is the horizontal distance

Horizontal Projection

If an object is thrown horizontally from a height, then the following are all true:

• Its path is initially horizontal but then curves steeper and steeper down
• The faster it is thrown then the further it will go before landing
• The time taken to land is independent

• Therefore with constant velocity in the x-component we have the displacement as Sx = UT
• If there is initially no vertical component (i.e. the object is thrown horizontally) then the y-displacement simplifies to: Sy = 1/2GT^2

Velocity Calculations

• The x velocity is a constant
• The y velocity undergoes constant acceleration (g) regardless of the initial conditions and independent of the x component Vy = GT

• Fy = FSinθ
• Fx = FCosθ