1. Constant Acceleration

2. 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 t v t v0v0 x0x0

3. 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 t a t a0a0 v0v0

4. Acceleration and Position  For constant acceleration the average acceleration equals the instantaneous acceleration.  Since the average of a line of constant slope is the midpoint: v t v0v0 ½t½t a 0 ( ½ t) + v 0

5. 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 = v 2 / 2ax = v 2 / 2a v 2 = 2 a xv 2 = 2 a x from

6. Accelerating a Mass  A loaded 747 jet has a mass of 4.1 x 10 5 kg and four engines.  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?  The distance and final velocity are used to get the acceleration.  The acceleration and mass give the force.

7. Pulley Acceleration  The normal force on m 1 equals the force of gravity.  The force of gravity is the only external force on m 2.  Both masses must accelerate together. m1m1 m2m2 FTFT F g = m 2 g  Consider two masses linked by a pulley m 2 is pulled by gravitym 2 is pulled by gravity m 1 is pulled by tensionm 1 is pulled by tension frictionless surfacefrictionless surface FTFT

8. Atwood’s Machine  In an Atwood machine both masses are pulled by gravity, but the force is unequal.  The heavy weight will move downward at (3.2 - 2.2 kg)(9.8 m/s 2 )/(3.2 + 2.2 kg) = 1.8 m/s 2.(3.2 - 2.2 kg)(9.8 m/s 2 )/(3.2 + 2.2 kg) = 1.8 m/s 2.  Using y = (1/2)at 2, it will take t 2 = 2(1.80 m)/(1.8 m/s 2 ) t = 1.4 s.t = 1.4 s. next