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 Describe motion in terms of frame of reference, displacement, time interval and velocity.  Calculate displacement, average velocity, time interval and acceleration.  Draw and interpret position vs. time and velocity vs. time graphs.

 The average velocity for any motion is › Where Δx is the displacement and Δt is the time interval.  The instantaneous velocity v is the velocity the object has at a particular time. › It is the average velocity over a very short time interval.

 If the velocity is constant, the instantaneous velocity is the average velocity.  v = v AV  The graph is a straight line.  The position is given by the equation

 Here the average velocity is not constant.  For the instantaneous velocity, take the average velocity over a very short time interval.  Graphically, this is the slope of the tangent line of the graph.

 When velocity changes, we have an acceleration.  Velocity can change in magnitude or direction.  Average acceleration is given by the formula:

vv aa vv aa vv aa vv aa vv  a = 0  v = 0  a or

 Accelerations can vary with time.  Many situations in physics can be modeled by a constant acceleration. › Constant acceleration means the object changes velocity at a constant rate.  When dealing with a constant acceleration situation, we will drop the subscript “AV”.

 a AV is the slope of the velocity vs. time graph.  If the velocity vs. time graph is a straight line, the acceleration is constant.  In this case, the formula for velocity is  t (s) v (m/s)

 Since and v is the height of the area under the velocity versus time graph, and t is the base of the velocity versus time graph, the area under a velocity versus time graph shows the displacement. ΔxΔx

 The displacement from time 0 to time t is the area under the velocity graph from 0 to t.  Area = ½ b h t (s) v (m/s)

 If the initial velocity is not zero, we have to include a rectangular piece.  Triangle Area = ½ b h  Rectangle = l x w t (s) v (m/s)

 If we don’t know v f, we can calculate it from a.  Area =l w + ½ b h t (s) v (m/s)

 Now we have derived three equations that apply to the motion with constant acceleration model

 If the motion begins at some other time other than t = 0, then we simply replace t with the time interval Δt.

 #1- An automobile with an initial speed of 4.3 m/s accelerates uniformly at the rate of 3.0 m/s 2. Find the final speed and the displacement after 5.0 s. v t Constant Acceleration Remember to list the GIVENS & UNKOWNS when setting up your equations!

v t

 #2 - A car starts from rest and travels for 5.0 s with a uniform acceleration of -1.5 m/s 2. What is the final velocity of the car? How far does the car travel in this time interval?

 All of the equations we have so far for this model involve time.  Sometimes, we are not told the time over which the motion occurs.  We can use two of these equations to eliminate time.

 A jet plane lands with a speed of 100 m/s and can accelerate uniformly at a maximum rate of -5.0 m/s 2 as it comes to rest. Can this airplane land at an airport where the runway is 0.80 km long?

 #3 Constant Acceleration