Constant AccelerationGraphs to FunctionsSlide 3Acceleration and PositionAcceleration RelationshipsAccelerating a MassPulley AccelerationAtwood’s MachineConstant AccelerationConstant AccelerationGraphs to FunctionsGraphs to FunctionsA 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.xtvtv0x000xtvx Constant AccelerationConstant AccelerationConstant 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 vtata0v000vtav Acceleration and PositionAcceleration and PositionFor 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:002000021)21( xtvtaxtvtax vtv0½ta0(½t) + v0Acceleration RelationshipsAcceleration RelationshipsAlgebra 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 00221xtvatx avvt0 0002000020221xavvvavvxxavvvavvax0vatv fromAccelerating a MassAccelerating a MassA 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/4 109.7N 109.7)m 107.1(2)m/s 81)(kg 101.4( 25553252engFFxmvmaFxva22axv 22Pulley AccelerationPulley AccelerationThe 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.m1m2FTFg = m2 gConsider 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 surfacegmmmaammgmmaFnet212212)(FTAtwood’s MachineAtwood’s MachineIn 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.gmmmmaammgmgmmaFnet21212121)(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
View Full Document
Unlocking...