A LEVEL MATHS - MECHANICS REVISION NOTES
1 KINEMATICS East
Distance - a scalar quantity with no direction
= 160 m
Displacement - a vector quantity – measured
from the starting position
= 40 m (East of starting point)
Position - a vector quantity – distance from a fixed origin
AVERAGE SPEED = 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
AVERAGE VELOCITY=
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
USING Position-Time and Velocity-Time GRAPHS
Position- time graph Velocity – time graph
stationary Moving back
Velocity ms-1
towards the
starting point
Gradient = velocity
Time (s)
VELOCITY TIME GRAPH
Gradient = acceleration
DISPLACEMENT – AREA
CONSTANT ACCELERATION UNDER THE GRAPH
(Straight Line)
VELOCITY = 0
Moving in the opposite
Velocity ms-1
direction – back towards the
starting position
1
, EQUATIONS FOR CONSTANT ACCELERATION -
s: displacement (m) u : initial velocity (ms-1) v : final velocity (ms-1) a : acceleration (ms-2)
t = time (s)
v = u + at v2 = u2 + 2as
s = ½(u + v)t s = ut + ½at2 s = vt - ½at2
Acceleration due to gravity is 9.8 ms-2 (unless given in the question)
Negative Acceleration means retardation/deceleration
You may need to show how the equations can be derived from the graph
A car starts from rest and reaches a speed of
15 ms-1 after travelling 25m with constant
A ball is thrown vertically upwards with a speed
acceleration. Assuming the acceleration remains
of 12 ms-1 from a height of 1.5 m. Calculate the
constant, how much further will the car travel
maximum height reached by the ball.
the next 4 seconds?
u = 12 ms-1
u = 0 ms v = 15 ms s = 25 m
-1 -1
a = -9.8 ms-2
At maximum height v = 0
v2 = u2 + 2as 152 = 2a × 25
v2 = u2 + 2as
a = 4.5 ms-2
0 = 144 - 2×9.8×s
s = 7.35 m
u = 15 ms-1 Maximum height = 1.5 + 7.35
t=4 = 8.85 m
a = 4.5 s = ut + ½ at2
s = 15 × 4 + ½ × 4.5 × 16
= 96 m
2 FORCES and ASSUMPTIONS
KEY FORCES
W : weight (mg = mass × 9.8)
R : reaction (normal reaction – at right angles to the point of contact)
F : friction (acts in a direction opposite to that in which the object is moving or is on the point
of moving)
T : Tension
ASSUMPTIONS
Objects are modelled as masses concentrated at a single point so no rotational forces.
Strings are inextensible (inelastic) so any stretch can be disregarded
Strings and rods are light (no mass) so weight can be disregarded
Pulleys are smooth so no frictional force at the pully needs to be considered.
2
1 KINEMATICS East
Distance - a scalar quantity with no direction
= 160 m
Displacement - a vector quantity – measured
from the starting position
= 40 m (East of starting point)
Position - a vector quantity – distance from a fixed origin
AVERAGE SPEED = 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
AVERAGE VELOCITY=
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
USING Position-Time and Velocity-Time GRAPHS
Position- time graph Velocity – time graph
stationary Moving back
Velocity ms-1
towards the
starting point
Gradient = velocity
Time (s)
VELOCITY TIME GRAPH
Gradient = acceleration
DISPLACEMENT – AREA
CONSTANT ACCELERATION UNDER THE GRAPH
(Straight Line)
VELOCITY = 0
Moving in the opposite
Velocity ms-1
direction – back towards the
starting position
1
, EQUATIONS FOR CONSTANT ACCELERATION -
s: displacement (m) u : initial velocity (ms-1) v : final velocity (ms-1) a : acceleration (ms-2)
t = time (s)
v = u + at v2 = u2 + 2as
s = ½(u + v)t s = ut + ½at2 s = vt - ½at2
Acceleration due to gravity is 9.8 ms-2 (unless given in the question)
Negative Acceleration means retardation/deceleration
You may need to show how the equations can be derived from the graph
A car starts from rest and reaches a speed of
15 ms-1 after travelling 25m with constant
A ball is thrown vertically upwards with a speed
acceleration. Assuming the acceleration remains
of 12 ms-1 from a height of 1.5 m. Calculate the
constant, how much further will the car travel
maximum height reached by the ball.
the next 4 seconds?
u = 12 ms-1
u = 0 ms v = 15 ms s = 25 m
-1 -1
a = -9.8 ms-2
At maximum height v = 0
v2 = u2 + 2as 152 = 2a × 25
v2 = u2 + 2as
a = 4.5 ms-2
0 = 144 - 2×9.8×s
s = 7.35 m
u = 15 ms-1 Maximum height = 1.5 + 7.35
t=4 = 8.85 m
a = 4.5 s = ut + ½ at2
s = 15 × 4 + ½ × 4.5 × 16
= 96 m
2 FORCES and ASSUMPTIONS
KEY FORCES
W : weight (mg = mass × 9.8)
R : reaction (normal reaction – at right angles to the point of contact)
F : friction (acts in a direction opposite to that in which the object is moving or is on the point
of moving)
T : Tension
ASSUMPTIONS
Objects are modelled as masses concentrated at a single point so no rotational forces.
Strings are inextensible (inelastic) so any stretch can be disregarded
Strings and rods are light (no mass) so weight can be disregarded
Pulleys are smooth so no frictional force at the pully needs to be considered.
2
