(1) The document discusses motion in one dimension, including displacement, velocity, acceleration, and free fall.
(2) Equations are provided relating displacement, velocity, acceleration, and time for motion with constant acceleration.
(3) An example problem is given about a dog and handler running towards each other with different initial velocities and accelerations to calculate their positions over time.
(1) The document discusses motion in one dimension, including displacement, velocity, acceleration, and free fall.
(2) Equations are provided relating displacement, velocity, acceleration, and time for motion with constant acceleration.
(3) An example problem is given about a dog and handler running towards each other with different initial velocities and accelerations to calculate their positions over time.
(1) The document discusses motion in one dimension, including displacement, velocity, acceleration, and free fall.
(2) Equations are provided relating displacement, velocity, acceleration, and time for motion with constant acceleration.
(3) An example problem is given about a dog and handler running towards each other with different initial velocities and accelerations to calculate their positions over time.
(1) The document discusses motion in one dimension, including displacement, velocity, acceleration, and free fall.
(2) Equations are provided relating displacement, velocity, acceleration, and time for motion with constant acceleration.
(3) An example problem is given about a dog and handler running towards each other with different initial velocities and accelerations to calculate their positions over time.
• The analysis of motion is separated into two distinct parts: (1) the description of the motion [kinematics] and (2) the cause of the motion [dynamics]. The discussion of kinematics is separated into two parts: motion in one spatial dimension and motion in two dimensions. Displacement and Distance In one dimension, the position of an object is described by the x coordinate. The goal of kinematics is to find the position as a function of time x = x(t). We define displacement as the vector from the initial position to the final position: ∆𝑥 = 𝑥𝑓 − 𝑥𝑖 … … … … … . . 2.1 Velocity and Speed The (average) velocity is defined by ∆𝑥 𝑥 − 𝑥0 𝑣= = … … … … … … 2.2 ∆𝑡 𝑡 − 𝑡0 where 𝑥0 is the position at time 𝑡0 : 𝑥0 = x(𝑡0 ). The unit of velocity is [v] = [x]/[t] = m/s. Using 𝑡0 = 0 and solve Eq.(2.2) for the coordinate x when the velocity is constant: x(t) = 𝑥0 + vt ………..2.3
The (average) speed is defined as the ratio of traveled distance divided by
the change in time: distance traveled speed = …………2.4 ∆𝑡 Example 1: A jaguar can reach speeds of 30 m/s. The fastest person is capable of reaching a speed of 10 m/s. Suppose the person and the jaguar are 500 m apart. Assume that they are both constantly at their top speed. How long does it take the jaguar to catch up to the person? • Acceleration In general, the velocity varies with time; mathematically, we say that the velocity is a function of time v = v(t). This is called instantaneous velocity. The rate of change in velocity is called acceleration: ∆𝑣 𝑣 − 𝑣0 𝑎= = … … … . . 2.5 ∆𝑡 𝑡 − 𝑡0 where v is the velocity at time t , and 𝑣𝑜 is the velocity at time 𝑡𝑜 . The unit of acceleration is [a] = [v]/[t] = m/s 2 . We set 𝑡0 = 0 , and find, 𝑣 = 𝑣𝑜 + 𝑎𝑡 (instantaneous velocity) …….2.6
The acceleration can be positive or negative. The object speeds up, when both velocity and acceleration are positive or negative; and the object slows down, when the velocity is positive and the acceleration is negative [or vice versa]. The average velocity during the time interval [0,t] follows, 1 1 𝑣 = 𝑣 + 𝑣𝑜 = 𝑣𝑜 + 𝑎𝑡, 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 … .2.7 2 2 Note that the RHS of Eq. (2.7) is equal to the instantaneous velocity at time t/2: v(t/2) = v0 + a(t/2). This is called the midpoint rule. The displacement follows, 1 𝑎 2 𝑥 − 𝑥𝑜 = 𝑣𝑡ҧ = 𝑣𝑜 + 𝑎𝑡 . 𝑡 = 𝑣𝑜 𝑡 + 𝑡 … … … .2.8 2 2 We find an equation that relates displacement to velocity by eliminating the time using Eq 2.6 t = (v − v0 )/a , so that ∆x = x − x0 = (v0 + v)/2 · (v − v0 )/a so that 𝑣 2 = 𝑣𝑜 2 + 2𝑎∆𝑥 … … … … … … … … … … … … … … . . 2.9 Eqs. (2.6)-(2.9) are the basis for motion with constant acceleration. We will always assume that the acceleration is piecewise constant, i.e., is constant during finite time intervals. • Exercises • A dog and her handler are at rest, and are facing each other at a distance of 40.0 m. On command, they run towards each other: the handler runs at a constant speed vp = 2.5 m/s, while the dog first runs with constant acceleration ad = 2.0 m/s 2 until she reaches her maximum speed of vd = 4.5 m/s. a) Calculate the time it takes the dog to reach her maximum speed. Choose t = 0 when the handler and dog start to run. b) Find the positions of the handler and dog at the instant the dog reaches her maximum speed. Choose the coordinate, such as that the handler is at the origin x = 0 at time t = 0. Free Fall • Free fall is an important case of motion with constant acceleration. All objects near the Earth’s surface fall with the same acceleration, if effects due to air resistance can be eliminated or ignored. • Experiment shows that if the effects of the air can be neglected, all bodies at a particular location fall with the same downward acceleration, regardless of their size or weight. • The idealized motion that results under all of these assumptions is called free fall, although it includes rising as well as falling motion. The constant acceleration of a freely falling body is called the acceleration due to gravity, and we denote its magnitude with the letter g = 9.8 m/s2 . • On the surface of the moon, the acceleration due to gravity is caused by the attractive force of the moon rather than the earth, and g = l.6 m/s2 . Near the surface of the sun, g = 270 m/s2 . • We choose the +𝑦 – axis upwards, so that 𝑣 > 0 𝑎𝑛𝑑 𝑣 < 0 describes an object flying upwards and downwards, respectively. • Then ay = −𝑔 [negative sign indicates downwards]. We then have the kinematics equations for free fall, Example 1 : A ball is dropped from the roof of a tower with a height of 81 m. a) How long will it take to hit the ground? b) Calculate the speed without calculating the time. Exercise : A ball is thrown along the vertical. The coordinate of the ball is observed as a function of time y(t). We choose y = 0 at the ground. The ball is at y = 13.2 m at time t = 0, and at y = 5.0 m at time t = 3.2 s. a) Find the displacement and average velocity of the ball between t = 0 and t = 3.2 s. b) Find the initial and final velocity of the ball. c) What is the height of the ball’s peak above ground? d) What is the average speed of the ball between t = 0 and t = 3.2 s?
