Transcript Constant Acceleration

Constant Acceleration
Graphs to Functions


A simple graph of constant
velocity corresponds to a
position graph that is a
straight line.
The functional form of the
position is
x  v0t  x0

This is a straight line and
only applies to straight lines.
x
t
x0
v
v0
t
Constant Acceleration



Constant velocity gives a
straight line position graph.
Constant acceleration gives
a straight line velocity graph.
The functional form of the
velocity is
v  a0t  v0
v
t
v0
a
a0
t
Acceleration and Position

For constant acceleration the
average acceleration equals
the instantaneous
acceleration.
v
a0(½t) + v0
v0

Since the average of a line
of constant slope is the
midpoint:
1
1 2
x  ( a0t  v0 )t  x0  a0t  v0t  x0
2
2
½t
t
Acceleration Relationships

Algebra can be used to
eliminate time from the
equation.

This gives a relation
between acceleration,
velocity and position.

For an initial or final velocity
of zero. This becomes
• x = v2 / 2a
• v2 = 2 a x
v  v0
from
a
1
x  at 2  v0t  x0
2
t
v  at  v0
1  v  v0 
 v  v0 
x  a

v

  x0
0
2  a 
 a 
2
2

v  v0   x
v  v0 
x
v
2a
0
a
0
Accelerating a Mass

A loaded 747 jet has a
mass of 4.1 x 105 kg
and four engines.

The distance and final velocity
are used to get the acceleration.
v  2ax
2
It takes a 1700 m
runway at constant
thrust (force) to reach a
takeoff speed of 81 m/s
(290 km/h).
What is the force per
engine?

v2
a
2x
The acceleration and mass give
the force.
mv 2
F  ma 
2x
(4.1105 kg )(81 m/s ) 2
5
F

7
.
9

10
N
3
2(1.7 10 m)
Feng  7.9 105 N/4  2.0 105 N
Pulley Acceleration

Consider two masses linked
by a pulley
• m2 is pulled by gravity
• m1 is pulled by tension
• frictionless surface



The normal force on m1
equals the force of gravity.
The force of gravity is the
only external force on m2.
Both masses must
accelerate together.
Fnet  ma
FT
m1
FT
m2
Fg = m2 g
m2 g  (m1  m2 )a
m2
a
g
m1  m2
Atwood’s Machine

In an Atwood machine both
masses are pulled by gravity,
but the force is unequal.
Fnet  ma
m1 g  m2 g  (m1  m2 )a
m1  m2
a
g
m1  m2

The heavy weight will move downward at
• (3.2 - 2.2 kg)(9.8 m/s2)/(3.2 + 2.2 kg) = 1.8 m/s2.

Using y = (1/2)at2, it will take t2 = 2(1.80 m)/(1.8 m/s2)
• t = 1.4 s.
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