constant acceleration
DESCRIPTION
Constant Acceleration. x. v. t. t. 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 This is a straight line and only applies to straight lines. x 0. v 0. v. a. t. t. - PowerPoint PPT PresentationTRANSCRIPT
Constant AccelerationConstant Acceleration
Graphs to FunctionsGraphs to Functions
A simple graph of constant A simple graph of constant velocity corresponds to a velocity corresponds to a position graph that is a position graph that is a straight line.straight line.
The functional form of the The functional form of the position is position is
This is a straight line and This is a straight line and only applies to straight lines.only applies to straight lines.
x
t
v
t
v0
x0
00 xtvx
Constant AccelerationConstant Acceleration
Constant velocity gives a Constant velocity gives a straight line position graph.straight line position graph.
Constant acceleration gives Constant acceleration gives a straight line velocity graph.a straight line velocity graph.
The functional form of the The functional form of the velocity is velocity is
v
t
a
t
a0
v0
00 vtav
Acceleration and PositionAcceleration and Position
For constant acceleration the For constant acceleration the average acceleration equals average acceleration equals the instantaneous the instantaneous acceleration.acceleration.
Since the average of a line Since the average of a line of constant slope is the of constant slope is the midpoint:midpoint:
002
0000 2
1)
2
1( xtvtaxtvtax
v
tv0
½t
a0(½t) + v0
Acceleration RelationshipsAcceleration Relationships
Algebra can be used to Algebra can be used to eliminate time from the eliminate time from the equation.equation.
This gives a relation This gives a relation between acceleration, between acceleration, velocity and position.velocity and position.
For an initial or final velocity For an initial or final velocity of zero. This becomesof zero. This becomes• xx = = vv22 / 2 / 2aa
• vv22 = 2 = 2 a xa x
002
2
1xtvatx
a
vvt 0
0
00
20
00
0
2
0
2
2
1
xa
vvv
a
vvx
xa
vvv
a
vvax
0vatv from
Accelerating a MassAccelerating a Mass
A loaded 747 jet has a A loaded 747 jet has a mass of 4.1 x 10mass of 4.1 x 1055 kg kg and four engines.and four engines.
It takes a 1700 m It takes a 1700 m runway at constant runway at constant thrust (force) to reach a thrust (force) to reach a takeoff speed of 81 m/s takeoff speed of 81 m/s (290 km/h).(290 km/h).
What is the force per What is the force per engine? engine?
The distance and final velocity The distance and final velocity are used to get the acceleration.are used to get the acceleration.
The acceleration and mass give The acceleration and mass give the force.the force.
N 100.2N/4 109.7
N 109.7)m 107.1(2
)m/s 81)(kg 101.4(
2
55
53
25
2
engF
F
x
mvmaF
x
va
2
2
axv 22
Pulley AccelerationPulley Acceleration
The normal force on The normal force on mm11
equals the force of gravity.equals the force of gravity. The force of gravity is the The force of gravity is the
only external force on only external force on mm22..
Both masses must Both masses must accelerate together.accelerate together.
m1
m2
FT
Fg = m2 g
Consider two masses linked Consider two masses linked by a pulleyby a pulley• mm22 is pulled by gravity is pulled by gravity
• mm11 is pulled by tension is pulled by tension
• frictionless surfacefrictionless surface
gmm
ma
ammgm
maFnet
21
2
212 )(
FT
Atwood’s MachineAtwood’s Machine
In an Atwood machine both In an Atwood machine both masses are pulled by gravity, masses are pulled by gravity, but the force is unequal.but the force is unequal.
gmm
mma
ammgmgm
maFnet
21
21
2121 )(
The heavy weight will move downward at The heavy weight will move downward at • (3.2 - 2.2 kg)(9.8 m/s(3.2 - 2.2 kg)(9.8 m/s22)/(3.2 + 2.2 kg) = 1.8 m/s)/(3.2 + 2.2 kg) = 1.8 m/s22..
Using Using yy = (1/2) = (1/2)atat22, it will take , it will take tt22 = 2(1.80 m)/(1.8 m/s = 2(1.80 m)/(1.8 m/s22))• tt = 1.4 s. = 1.4 s. next