Presentation on theme: "Motion with Constant Acceleration"— Presentation transcript:
1 Motion with Constant Acceleration McNutt Physics – 09/16/2013
2 The Story so far…. 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.
3 Position vs. Time for Constant Velocity Motion If the velocity is constant, the instantaneous velocity is the average velocity.v = vAVThe graph is a straight line.The position is given by the equation
4 Position vs. Time for Accelerated Motion 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.
5 Acceleration When velocity changes, we have an acceleration. Velocity can change in magnitude or direction.Average acceleration is given by the formula:
6 Accelerations can be positive or negative in 1-d motion. a or
7 Constant Acceleration Model 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”.
8 Velocity vs. time for constant acceleration aAV 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 isv(m/s)t (s)
9 Displacement on a Velocity vs. Time graph 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.Δx
10 Displacement for constant acceleration The displacement from time 0 to time t is the area under the velocity graph from 0 to t.Area = ½ b hv(m/s)t (s)
11 Displacement for constant acceleration If the initial velocity is not zero, we have to include a rectangular piece.Triangle Area = ½ b hRectangle = l x wv(m/s)t (s)
12 Displacement for constant acceleration If we don’t know vf, we can calculate it from a.Area =l w + ½ b hv(m/s)t (s)
13 Equations of Motion for Constant Acceleration Now we have derived three equations that apply to the motion with constant acceleration model
14 Formulas for other time intervals If the motion begins at some other time other than t = 0, then we simply replace t with the time interval Δt.
15 Practice 2D, p. 55#2- An automobile with an initial speed of 4.3 m/s accelerates uniformly at the rate of 3.0 m/s2. Find the final speed and the displacement after 5.0 s.vtConstant Acceleration
17 Practice 2D, p. 55#3- A car starts from rest and travels for 5.0 s with a uniform acceleration of -1.5 m/s2. What is the final velocity of the car? How far does the car travel in this time interval?
18 One Other Equation for Constant Acceleration 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.
20 Practice 2C, p. 53A jet plane lands with a speed of 100 m/s and can accelerate uniformly at a maximum rate of -5.0 m/s2 as it comes to rest. Can this airplane land at an airport where the runway is 0.80 km long?