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Displacement
–
time graphs
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Acceleration
–
a change in velocity
Velocity changes when there is a change in its
magnitude
(i.e. a change in speed), a change in its
direction
, or both.
So acceleration
can include:
speeding up
slowing down (
deceleration
)
changing direction (e.g.
centripetal acceleration
)
So even though a geostationary satellite is travelling in a
circle at a steady speed, it is actually
accelerating
as it
constantly changes direction!
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Velocity
–
time graphs
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Four ‘
suvat
’ equations
Motion under constant acceleration can be described using
the following four equations:
1.
v
=
u
+
at
2.
s
=
ut
+ ½
at
2
3.
v
2
=
u
2
+ 2
as
4.
s
= ½(
u
+
v
)
t
These are known as the ‘
suvat
’ or
constant acceleration
equations
, where
u
is the initial velocity,
a
is the
acceleration, and
s
and
v
are the displacement and velocity
at time
t
. How can these equations be derived?
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Using the
suvat
equations
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Analysing a velocity
–
time graph
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Acceleration under gravity
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Acceleration of freefall
An object that falls to the ground with no forces acting on it
except gravity is said to be in
freefall
.
This can only occur when the effects
of air resistance are
negligible
.
Any object in freefall,
close to the
Earth’s surface, experiences vertical
acceleration of
9.81
ms

2
downwards.
This is often denoted by the letter
g
.
‘Freefall’ includes both ‘rising’
and ‘falling’ motion, whether a
projectile follows a
parabola
or
a simple vertical line.
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Capturing projectile motion
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Equations of projectile motion
An object in freefall:
moves at a constant horizontal (
x
) velocity
moves at a constant vertical (
y
) acceleration.
a
x
= 0
a
y
=
g
The following equations can therefore be applied.
Can you see how they have been derived?
x
=
v
x
t
y
=
v
y
=
u
y
+ g
t
y
=
u
y
t
+
½
g
t
2
v
y
2
=
u
y
2
+ 2
gy
u
y
+
v
y
t
constant
x
velocity
suvat
equations
for
u
y
and
v
y
with
a
=
g
2
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Birdman rally
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Height of a projectile
A tennis player hits a volley just above ground level, in a
direction perpendicular to the net.
The ball leaves her racquet at 8.2
ms

1
at an angle of 34
°
to the horizontal.
Will the ball clear the net if it is 2.3
m
away and 95
cm high at this point?
What assumptions should you
make to solve this problem?
no air resistance
no spin
initial height is zero.
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Height of a projectile
We need to calculate the value of
y
at
x
= 2.3
m and
determine whether or not it is greater than 0.95
m.
What are the relevant
equations of motion?
x
=
v
X
t
y
=
u
y
t
+
½
g
t
2
First, use the
x
equation to calculate
t
when
x
is 2.3.
2.3 = 8.2
×
cos34
°
×
t
t
= 0.34
s
34
°
8.2
ms

1
0.95
m
2.3
m
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Height of a projectile
Now substitute this value
of
t
into the
y
equation to
find
y,
and determine
whether or not it is
greater than 0.95
m.
y
= ((8.2
×
sin34
°
)
×
0.34) + (
½
×

9.81
×
0.34
2
)
So
y
is greater than 0.95 and the ball clears the net!
35
°
8.2
ms

1
0.95
m
2.3
m
So the ball reaches
x
= 2.3
m when
t
= 0.34
s.
y
=
u
y
t
+
½
g
t
2
y
= 0.99 m
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Interactive cannon
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Glossary
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What’s the keyword?
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Multiple
–
choice quiz
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