Acceleration. Graphs to Functions  A simple graph of constant velocity corresponds to a position graph that is a straight line.  The functional form.

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Presentation transcript:

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  This is a straight line and only applies to straight lines. x t v t v0v0 x0x0

Changing Velocity  In more complicated motion the velocity is not constant.  We can express a time rate of change for velocity just as for position,  v = v 2 - v 1.  The acceleration is the time rate of change of velocity: a =  v /  t.

Average Acceleration Example problem  A jet plane has a takeoff speed of 250 km/h. If the plane starts from rest, and lifts off in 1.2 min what is the average acceleration? a =  v /  t = [(250 km/h) / (1.2 min)] * (60 min/h) a =  v /  t = [(250 km/h) / (1.2 min)] * (60 min/h) a = 1.25 x 10 4 km/h 2 a = 1.25 x 10 4 km/h 2  Why is this so large? Is it reasonable?  Does the jet accelerate for an hour?

Instantaneous Acceleration  Instantaneous velocity is defined by a derivative.  Instantaneous acceleration is also defined by a derivative. v t P1P1 P2P2 P3P3 P4P4

Second Derivative  The acceleration is the derivative of velocity with respect to time.  The velocity is the derivative of position with respect to time.  This makes the acceleration the second derivative of position with respect to time.

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

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

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

Take Off Example problem  A jet plane has a takeoff speed of 250 km/h. If the plane starts from rest, and has a constant acceleration of 1.25 x 10 4 km/h 2, what is the length of the runway? x = v 2 / 2a = (250 km/h) 2 / (2.5 x 10 4 km/h 2 ) x = v 2 / 2a = (250 km/h) 2 / (2.5 x 10 4 km/h 2 ) x = 2.5 km x = 2.5 km  Is this reasonable? next