# Kinematics_1

Μηχανική

14 Νοε 2013 (πριν από 4 χρόνια και 6 μήνες)

117 εμφανίσεις

1

of 22

2

of 22

3

of 22

Displacement

time graphs

4

of 22

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!

5

of 22

Velocity

time graphs

6

of 22

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?

7

of 22

Using the
suvat

equations

8

of 22

Analysing a velocity

time graph

9

of 22

10

of 22

Acceleration under gravity

11

of 22

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.

12

of 22

Capturing projectile motion

13

of 22

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

14

of 22

Birdman rally

15

of 22

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.

16

of 22

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

17

of 22

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

18

of 22

Interactive cannon

19

of 22

20

of 22

Glossary

21

of 22