VCE Physics.comEquations of motion -
Equations of motion
1
• Displacement, velocity & acceleration
• Velocity-time graph
• The five kinematics equations
• Solving kinematics problems
• Gravity
VCE Physics.comEquations of motion -
• 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.
Displacement, velocity & acceleration
2
x = displacement (m)t = time (s)
a = acceleration (m/s2)u = initial velocity (m/s)v = final velocity (m/s)
VCE Physics.comEquations of motion -
Velocity-time graphs
3
• The gradient of the graph gives the acceleration.
• The area under the graph gives the displacement.
Velocity (m/s)
Time (s)
u
v
t
Acceleration
Displacement1 m/s x 1 s = 1m
VCE Physics.comEquations of motion -
Defining the equations (1)
4
• The gradient of the graph gives the acceleration.
Velocity (m/s)
Time (s)
u
v
t
a=
v-ut
v - u
t
VCE Physics.comEquations of motion -
Defining the equations (2)
5
• The area under the graph gives the displacement.
Velocity (m/s)
Time (s)
u
v
x =
12
(u +v )tu
v
t
Trapezium area
=12
(u +v )t
VCE Physics.comEquations of motion -
Defining the equations (3)
6
• The area under the graph gives the displacement.
Velocity (m/s)
Time (s)
u
v
t
v - u = at
ut
12
at2
Triangle area
=12
t × at
=12
at2
x =ut+
12
at2
VCE Physics.comEquations of motion -
Defining the equations (4)
7
• The area under the graph gives the displacement.
Velocity (m/s)
Time (s)
u
v
t
12
at2
x = vt-
12
at2
vt
VCE Physics.comEquations of motion -
Defining the equations (5)
8
• We can also define a rule that is not dependent on time.
x =
12
(u +v )t
a=
v-ut
t =v-u
a
x =
(u +v )2
(v −u)a
x =
v 2 −u2
2a
VCE Physics.comEquations of motion -
• A car accelerates from the lights, 0 to 60 km/h in 4.0 seconds
• Find the distance covered in that time.
Solving kinematics problems (1)
9
t = 4.0 s
Known information:
u = 0 m/s
v =
60 km/h3.6
= 17 m/s
x =
12
(u +v )t
x =
12
(0 + 17 m/s) × 4 s
x =34 m
(This is an average speed of 8.5 m/s over the 4 seconds.)
VCE Physics.comEquations of motion -
• 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?
Solving kinematics problems (2)
10
a = -6 m/s2
Known information:
v = 0 m/s
u =
100 km/h3.6
= 28 m/s
x =65 m
x =
v2 - u2
2a
x = (0 m/s)2 - (28 m/s)2
2 × -6 m/s2
(Stopping distance varies with the square of speed.
2 x speed = 4 x stopping distance.)
VCE Physics.comEquations of motion -
• 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?
Solving kinematics problems (3)
11
a = -10 m/s2
Known information:
v = -15 m/s
u = 15 m/s
If the ball lands at the height that it was hit from:
a=
v-ut
t =
- 15 m/s - 15 m/s-10 m/s2
t =
v-ua
t =3 sMaximum height at 1.5 s, v = 0
x =
12
(u +v )t
Acceleration due to gravityg ≈10 m/s2
x =
12
(15 m/s + 0 m/s)(1.5 s)
x = 11.25 m
VCE Physics.comEquations of motion -
• Objects falling (or rising up) experience an acceleration due to the force of gravity.
• g (on Earth) = 9.81 m/s2 ≈10 m/s2
• (Presuming that air resistance is insignificant: the projectile is heavy, small & not moving very fast.)
Gravity
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v = at
Time (s) Speed Distance
0 0 m/s 0 m
1 10 m/s 5 m
2 20 m/s 20 m
3 30 m/s 45 m
4 40 m/s 80 m
5 50 m/s 125 m
x =
12
at2
u = 0 m / s
0 m5 m
20 m
45 m
80 m
125 m
VCE Physics.comEquations of motion -
Graphs of motion under gravity
13
v = u + at
x =ut +
12
at2
u = 0 m / s
a = -10 m/s2
v = −10t
a = constant
Velocity
time (s)
Acceleration
time (s)
x = −5t2