Chapter 2 Motion Along a Line

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/20102 Motion Along a Line Position & Displacement Speed & Velocity Acceleration Describing motion in 1D Free Fall

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/20103 Introduction Kinematics - Concepts needed to describe motion - displacement, velocity & acceleration. Dynamics - Deals with the effect of forces on motion. Mechanics - Kinematics + Dynamics

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/20104 Goals of Chapter 2 Develop an understanding of kinematics that comprehends the interrelationships among physical intuition equations graphical representations When we finish this chapter you should be able to move easily among these different aspects.

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/20105 Kinematic Quantities Overview The words speed and velocity are used interchangably in everyday conversation but they have distinct meanings in the physics world.

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/20106 The position (x) of an object describes its location relative to some origin or other reference point. Position & Displacement 0 x2x2 0 x1x1 The position of the red ball differs in the two shown coordinate systems.

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ x (cm) 2 1 22 11 The position of the ball is The + indicates the direction to the right of the origin.

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ x (cm) 2 1 22 11 The position of the ball is The  indicates the direction to the left of the origin.

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/20109 The displacement is the change in an object’s position. It depends only on the beginning and ending positions. All Δ quantities will have the final value 1st and the inital value last.

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ x (cm) 2 1 22 11 Example: A ball is initially at x = +2 cm and is moved to x = -2 cm. What is the displacement of the ball?

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ Example: At 3 PM a car is located 20 km south of its starting point. One hour later its is 96 km farther south. After two more hours it is 12 km south of the original starting point. (a) What is the displacement of the car between 3 PM and 6 PM? x i = –20 km and x f = –12 km Use a coordinate system where north is positive.

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ (b) What is the displacement of the car from the starting point to the location at 4 pm? (c) What is the displacement of the car from 4 PM to 6 PM? Example continued x i = 0 km and x f = –96 km x i = –96 km and x f = –12 km

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ Velocity is a vector that measures how fast and in what direction something moves. Speed is the magnitude of the velocity. It is a scalar. Velocity: Rate of Change of Position

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ In 1-dimension the average velocity is v av is the constant speed and direction that results in the same displacement in a given time interval.

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ On a graph of position versus time, the average velocity is represented by the slope of a chord. x (m) t (sec) t1t1 t2t2 x1x1 x2x2

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ This is represented by the slope of a line tangent to the curve on the graph of an object’s position versus time. x (m) t (sec)

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ The area under a velocity versus time graph (between the curve and the time axis) gives the displacement in a given interval of time. v x (m/s) t (sec)

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ Example (text problem 2.11): Speedometer readings are obtained and graphed as a car comes to a stop along a straight- line path. How far does the car move between t = 0 and t = 16 seconds? Since there is not a reversal of direction, the area between the curve and the time axis will represent the distance traveled.

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ Example continued: The rectangular portion has an area of Lw = (20 m/s)(4 s) = 80 m. The triangular portion has an area of ½bh = ½(8 s) (20 m/s) = 80 m. Thus, the total area is 160 m. This is the distance traveled by the car.

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ The Most Important Graph- V vs T Area under the curve gives DISTANCE. The slope of the curve gives the ACCELERATION. The values of the curve gives the instantaneous VELOCITY. Negative areas are possible.

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ Acceleration: Rate of Change of Velocity These have interpretation s similar to v av and v.

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ Example (text problem 2.29): The graph shows speedometer readings as a car comes to a stop. What is the magnitude of the acceleration at t = 7.0 s? The slope of the graph at t = 7.0 sec is

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ Motion Along a Line With Constant Acceleration For constant acceleration the kinematic equations are: Also:

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ A Modified Set of Equations For constant acceleration the kinematic equations are: Also:

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ Visualizing Motion Along a Line with Constant Acceleration Motion diagrams for three carts:

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ Graphs of x, v x, a x for each of the three carts

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ A stone is dropped from the edge of a cliff; if air resistance can be ignored, we say the stone is in free fall. The magnitude of the acceleration of the stone is a free fall = g = 9.80 m/s 2, this acceleration is always directed toward the Earth. The velocity of the stone changes by 9.8 m/s every sec. Free Fall

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ Free Fall Assumption: acceleration due to gravity is g g = 9.8 m/s 2 ≈ 10 m/s 2

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ Example: You throw a ball into the air with speed 15.0 m/s; how high does the ball rise? Given: v iy = m/s; a y =  9.8 m/s 2 x y v iy ayay To calculate the final height, we need to know the time of flight. Time of flight from:

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ The ball rises until v fy = 0. The height: Example continued:

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ Example (text problem 2.45): A penny is dropped from the observation deck of the Empire State Building 369 m above the ground. With what velocity does it strike the ground? Ignore air resistance. 369 m x y Given: v iy = 0 m/s, a y =  9.8 m/s 2,  y =  369 m Unknown: v fy Use: ayay

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ How long does it take for the penny to strike the ground? (downward) Example continued: Given: v iy = 0 m/s, a y =  9.8 m/s 2,  y =  369 m Unknown:  t

MFMcGraw- PHY 1410Ch_02b-Revised 5/31/ Summary Position Displacement Versus Distance Velocity Versus Speed Acceleration Instantaneous Values Versus Average Values The Kinematic Equations Graphical Representations of Motion Free Fall