Motion Along a Straight Line (Motion in One-Dimension)

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1 Chapter 2 Motion Along a Straight Line (Motion in One-Dimension) Learn the concepts of displacement, velocity, and acceleration in one-dimension. Describe motions at constant acceleration. Be able to graph and interpret graphs as they describe motion. Be able to reason proportionally. Examine the special case of freely falling bodies. Consider relative motion.

2 The study of motion is divided into two areas: Kinematics This will be our focus in Chapters 2 (1-dimension) and 3 (2-dimensions). Kinematics describes the movement of the object, regarding its position, velocity, and acceleration. A coordinate system needs to be set up in order to describe its position, velocity, and acceleration. Dynamics Will come in Chapter 4 and after Dynamics answers the question "Why is this object moving? It is about how the application of forces affects the motion of an object.

3 2.1 Displacement and Average Velocity Coordinate system in 1-dimension: a single axis with a direction and an origin Example: x-axis or y-axis Displacement describes the change in position: It is a vector quantity pointing from the initial position (x 1 ) to the final position (x 2 ). Its magnitude (the distance) is equal to the length of the straight line connecting these two positions. In 1-dimension, the direction of a vector is either + or -, depending on if it is pointing in the positive direction of the axis or not. Units in SI unit system: meters (m) Average velocity: The displacement divided by the elapsed time Dt = t 2 t 1 It is also a vector. Direction in 1-dimension: either + or -.

4 Displacement and Average Velocity

5 Displacement and Average Velocity

6 Understand the Meaning of Average Velocity Graphically The average velocity between two positions is the slope of the line connecting these two corresponding points on a graph of position as a function of time.

7 2.2 Instantaneous Velocity How do we calculate the velocity at an instant of time? Instantaneous Velocity:

8 Understanding the direction and the magnitude of the instantaneous velocity graphically

9 2.3 Average and Instantaneous Acceleration When velocity changes with time acceleration Average Acceleration a av,x = v 2x v 1x t 2 t 1 = v x t Instantaneous Acceleration a x = lim t 0 v x t Units: m/s 2

10 Understand the Direction of Acceleration

11 Direction of Velocity and Acceleration in a Free Fall Problem Coordinate Free fall from rest Tossing a Ball upward y at the top v = 0 a = m/s 2 at the top v = 0 a = m/s 2 the way down v is negative a = m/s 2 the way up v is positive a = m/s 2 the way down v is negative a = m/s 2 0 ground

12 Example 2.4 Average and instantaneous acceleration, p. 37 The velocity of the object is given: v x (t) = (60.0 m/s) + (0.500 m/s 3 )t 2 (a) Find the change in velocity in the time interval between t 1 = 1.00 s and t 2 = 3.00 s. Dv x = v x (t 2 ) v x (t 1 ) = v x (3.00 s) v x (1.00 s) (b) Find the average acceleration during this time interval. a av,x = Dv x /Dt, where Dt = t 2 t 1 = (3.00 s) (1.00 s) = 2.00 s (c) Estimate the instantaneous acceleration at time t 1 = 1.00 s by taking Dt = 0.10 s. a av,x = Dv x /Dt = [v x (1.10 s) v x (1.00 s)]/(0.10 s)

13 2.4 Motion with Constant Acceleration Main goal: Understand the three (+1) kinematic equations and use them to solve problems Question: How can we predict the motion of a particle? We must at least know where it starts at (x 0 ) and what its initial velocity (v 0x ) is. These are called the initial conditions (at time t = 0, or, any given time) in a given coordinate (x-axis). Now Consider: a particle that is moving with a constant acceleration, a x = constant. Question: What is the velocity of the particle at any other time t? Since the acceleration is defined by a x = v x t v x (0) t 0 = v x t v 0x, t the velocity of the particle a time t is v x t = v 0x + a x t...(2.6) This is the first of the three kinematic equations.

14 Question: How can we predict the position of the particle at any time t? Since the average velocity is defined by v av,x = x t x(0) t 0 we have x t = x 0 + v av,x t. But, what is v av,x? Since v x t is linear in time t, v x t = v 0x + a x t, = x t x 0, t we have v av,x = 1 2 [v x t + v 0x ]. Therefore, x t = x [v x t + v 0x ]t = x [ v 0x + a x t + v 0x ]t, or, x t = x 0 + v 0x t a xt 2 (2.10) This is the second of the three kinematic equations.

15 Question: What if you encounter a situation in which time is neither given nor asked of? Here comes the third of the three kinematic equations, which relates to each other the position, velocity, and acceleration of a particle. Let us consider the first two kinematic equations: v x t = v 0x + a x t.....(2.6) x t = x 0 + v 0x t a xt 2.. (2.10) The goal is to use substitution to eliminate time t. Solve for time in terms of other quantities using (2.6): and then substitute into (2.10) to eliminate time: x = x 0 + v 0x v x v 0x a x + a x 2 v x v 0x a x t = v x v 0x a x, 2 = x 0 + v x 2 v2 0x, 2a x or, v 2 x = v 2 0x + 2a x (x x 0 ) (2.11) This is the third of the kinematic equations.

16 v x t = v 0x + a x t...(2.6) Application of Kinematic Equations x t = x 0 + v 0x t a xt 2 (2.10) v 2 x = v 2 0x + 2a x x x 0 (2.11) Example 2.9 on page 45 v av,x = 1 2 [v x t + v 0x ].. (2.7) (a) Find the time it takes for the officer to catch up with the motorist? The approach: Find the expressions as functions of time for the positions of the police and the motorist, and then, set these two to equal to each other to solve for time. Position of the Officer: Position of the motorist: x P t = a P x C t = 0 + v C0 t Let x P t = x C t to solve for t: a P 2 t2 = v C0 t, which is the same as or t( a P t v 2 C0) = 0, t = 2v C0 = 10 s. a P 2 t2 (b) What is the speed of the officer at that time? 2v C0 v P t = 0 + a P = a P a P = 2v C0 = 30 m/s (c) The distance the office has traveled at that point, x P t = a P 2 t2 = a P 2 (2v C0 a P ) 2 = 2v C0 2 a P = 150 m

17 2.5 Proportional Reasoning Pay attention to the functional dependence, rather than focusing on numerical numbers. An object, initially at rest, starts to accelerate at t = 0 with a constant acceleration. If it gains a speed 10 m/s after a certain time, what is its speed when the time is doubled? Equation to consider: v x t = v 0x + a x t = a x t with v 0x = 0. An object, initially at rest, starts to accelerate at t = 0 with a constant acceleration. If it travels a distance 10 m in the first 5 s, what is the distance it travels in the first 10 s? Equation to consider: x t = x 0 + v 0x t a xt 2 with x 0 = 0 and v 0x = 0. If the gravitational potential energy between a star and a planet is V when their distance is R, what should be the gravitational potential energy if their distance is cut in half? Equation to consider: V = G Mm R If the gravitational attraction a star exerts on a planet is F when their distance is R, what should be the gravitational attraction if their distance is doubled? Equation to consider: F = G Mm R 2

2.6 Freely Falling Objects v y t = v 0y + a y t...(2.6) y t = y 0 + v 0y t + 1 2 a yt 2.. (2.10) v 2 y = v 2 0y + 2a y y y 0 (2.11) Example 2.

18 2.6 Freely Falling Objects v y t = v 0y + a y t...(2.6) y t = y 0 + v 0y t a yt 2.. (2.10) v 2 y = v 2 0y + 2a y y y 0 (2.11) Example 2.10 on page 48 With the given y-axis and initial conditions, calculate the position using eqn.(2.10) and the velocity using eqn.(2.6). Pay attention to the sign of each quantity. v av,y = 1 2 [v y t + v 0y ] (2.7) a y = g = 9.8 m/s 2 if the y-axis points up v y t = v 0y gt...(2.6) y t = y 0 + v 0y t 1 2 gt2 (2.10) v 2 y = v 2 0y 2g y y 0..(2.11) v av,y = 1 2 [v y t + v 0y ].. (2.7)

19 v y t = v 0y gt...(2.6) y t = y 0 + v 0y t 1 2 gt2. (2.10) v 2 y = v 2 0y 2g y y 0....(2.11) v av,y = 1 [v 2 y t + v 0y ]...(2.7) Example 2.11 A ball on the roof on p Given: y 0 = 0 and v 0y = 15.0 m/s Find: (a) y and v y at t = 1.00 s, and, y and v y at t = 4.00 s (b) v y when y = 5.00 m (c) maximum height y max and the time reaching y max (a) Use eqn.(2.10) for y and use eqn.(2.6) for v y. (b) There are more than one ways for calculating v y at y = 5.00 m. (b1) Use eqn.(2.11), with v 0y = 15.0 m/s and (y y 0 ) = 5.00 m. You get two answers when you take the square-root: v y = ± 11.3 m/s, corresponding to the velocities on the ways up and down. (b2) A less convenient way is to use eqn.(2.10) to solve for the times at which the ball crosses y = 5.00 m, on the ways up and down. Then, use eqn.(2.6) to calculate the corresponding velocities.

20 v y t = v 0y gt....(2.6) y t = y 0 + v 0y t 1 2 gt2 (2.10) v 2 y = v 2 0y 2g y y 0...(2.11) v av,y = 1 [v 2 y t + v 0y ]..(2.7) Example 2.11 A ball on the roof on p Given: y 0 = 0 and v 0y = 15.0 m/s Find: (a) y and v y at t = 1.00 s, and, y and v y at t = 4.00 s (b) v y when y = 5.00 m (c) maximum height y max and the time reaching y max (c) Meaning of at the maximum height? v y = 0. So, you are asked to find the y max and t max for which v y = 0. There are more than one ways. (c1) Use eqn.(2.11) to find y max, with y 0 = 0, v y = 0, and v 0y = 15.0 m/s. Then, use eqn.(2.10) to find the time t max that it reaches the top. (c2) use eqn.(2.6) to find the time t max that it reaches the top. Then, use eqn.(2.10) to calculate y max. (c3) Once you got t max, use eqn.(2.7) to calculate y max y max = v av,y t max = [(v y t + v 0y )/2] t max

21 2.7 Relative Velocity Along a Straight Line (in One-Dimension) Positions are relative: Velocities are also relative: x W/C = x W/T + x T/C v W/C = v W/T + v T/C Pay attention to the sign: position and velocity are vectors.

Chapter 2. Motion in One Dimension

Chapter 2. Motion in One Dimension Chapter 2 Motion in One Dimension Types of Motion Translational An example is a car traveling on a highway. Rotational An example is the Earth s spin on its axis. Vibrational An example is the back-and-forth