Kinematics_1

doutfanaticalΜηχανική

14 Νοε 2013 (πριν από 3 χρόνια και 6 μήνες)

84 εμφανίσεις

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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