1INTRODUCTION & RECTILINEAR KINEMATICS: CONTINUOUS MOTION (Sections 12 Today’s Objectives:Students will be able to find the kinematic quantities (position, displacement, velocity, and acceleration) of a particle traveling along a straight path.In-Class Activities:• Check homework, if any• Reading quiz• Applications• Relations between s(t), v(t), and a(t) for generalrectilinear motion• Relations between s(t), v(t),and a(t) when acceleration is constant• Concept quiz• Group problem solving• Attention quiz
2An Overview of Mechanics Mechanics: the study of how bodies react to forces acting on themStatics: the study of bodies in equilibriumDynamics:1. Kinematics – concerned with the geometric aspects of motion2. Kinetics - concerned with the forces causing the motion
3Dynamics Dynamics includes: Kinematics: study of the geometry of motion. Kinematics is used to relate displacement, velocity, acceleration, and time without reference to the cause of motion.Geometry of motion or Calculus of motionKinetics: study of the relations existing between the forces acting on a body, the mass of the body, and the motion of the body. Kinetics is used to predict the motion caused by given forces or to determine the forces required to produce a given motion.Newtons Second LawEnergy MethodsImpulse-Momentum
4Particle:A body without size or shape that will be assumed to occupy a single point in space.This greatly simplifies mechanics problems and when considered appropriately, gives reasonable first results.Assumption: NO ROTATIONCaution: Can be a bit unrealisticLater on we will look at Rigid Body MotionKinematics-geometry of motionKinetics-Newtons Second Law (+ some Kinematics)Types of MotionRectilinear motion: position, velocity, and acceleration of a particle as it moves along a straight line.Curvilinear motion: position, velocity, and acceleration of a particle as it moves along a curved line in two or three dimensions.
5APPLICATIONSThe motion of large objects, such as rockets, airplanes, or cars, can often be analyzed as if they were particles.Why?If we measure the altitude of this rocket as a function of time, how can we determine its velocity and acceleration?
6APPLICATIONS (continued) A train travels along a straight length of track.Can we treat the train as a particle?If the train accelerates at a constant rate, how can we determine its position and velocity at some instant?
7POSITION AND DISPLACEMENT A particle travels along a straight-line path defined by the coordinate axis s.The position of the particle at any instant, relative to the origin, O, is defined by the position vector r, or the scalar s. Scalar s can be positive or negative. Typical units for r and s are meters (m) or feet (ft).The displacement of the particle is defined as its change in position.Vector form: Δ r = r’ - rScalar form: Δ s = s’ - sThe total distance traveled by the particle, sT, is a positive scalar that represents the total length of the path over which the particle travels.
8The instantaneous velocity is the time-derivative of position. Velocity is a measure of the rate of change in the position of a particle. It is a vector quantity (it has both magnitude and direction). The magnitude of the velocity is called speed, with units of m/s or ft/s.The average velocity of a particle during a time interval Δt isvavg = Δr/Δt = Δr/Δt iThe instantaneous velocity is the time-derivative of position.v = dr/dtSpeed is the magnitude of velocity: v = ds/dtAverage speed is the total distance traveled divided by elapsed time: (vsp)avg = sT/ Δ t
9Rectilinear Motion: Position, Velocity & Acceleration Particle moving along a straight line is said to be in rectilinear motion.Position coordinate of a particle is defined by positive or negative distance of particle from a fixed origin on the line.The motion of a particle is known if the position coordinate for particle is known for every value of time t. Motion of the particle may be expressed in the form of a function, e.g.,or in the form of a graph x vs. t.
10Consider particle which occupies position P at time t and P’ at t+Δt, Average velocityInstantaneous velocityInstantaneous velocity may be positive or negative. Magnitude of velocity is referred to as particle speed.From the definition of a derivative,e.g.,
11ACCELERATIONAcceleration is the rate of change in the velocity of a particle. It is a vector quantity. Typical units are m/s2 or ft/s2.The instantaneous acceleration is the time derivative of velocity.Vector form: a = dv/dtScalar form: a = dv/dt = d2s/dt2Acceleration can be positive (speed increasing) or negative (speed decreasing).As the book indicates, the derivative equations for velocity and acceleration can be manipulated to geta ds = v dv
12Consider particle with velocity v at time t and v’ at t+Δt, Instantaneous accelerationInstantaneous acceleration may be:positive: increasing positive velocityor decreasing negative velocitynegative: decreasing positive velocityor increasing negative velocity.From the definition of a derivative,
13Consider particle with motion given by at t = 0, x = 0, v = 0, a = 12 m/s2at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2at t = 6 s, x = 0, v = -36 m/s, a = 24 m/s2
14Recall, motion of a particle is known if position is known for all time t. Typically, conditions of motion are specified by the type of acceleration experienced by the particle. Determination of velocity and position requires two successive integrations.Three classes of motion may be defined for:acceleration given as a function of time, a = f(t)- acceleration given as a function of position, a = f(x)- acceleration given as a function of velocity, a = f(v)
15SUMMARY OF KINEMATIC RELATIONS: RECTILINEAR MOTION• Differentiate position to get velocity and acceleration.v = ds/dt ; a = dv/dt or a = v dv/ds• Integrate acceleration for velocity and position.Velocity:ò=tovdtadvsdsorò=tosdtvdsPosition:• Note that so and vo represent the initial position and velocity of the particle at t = 0.
16CONSTANT ACCELERATION The three kinematic equations can be integrated for the special case when acceleration is constant (a = ac) to obtain very useful equations. A common example of constant acceleration is gravity; i.e., a body freely falling toward earth. In this case, ac = g = 9.81 m/s2 = 32.2 ft/s2 downward. These equations are:tavco+=yields2(1/2)as)-(s2a(v
17Acceleration given as a function of time, a = f(t): Acceleration given as a function of position, a = f(x):
18Acceleration given as a function of velocity, a = f(v):
19EXAMPLEGiven: A motorcyclist travels along a straight road at a speed of 27 m/s. When the brakes are applied, the motorcycle decelerates at a rate of -6t m/s2.Find: The distance the motorcycle travels before it stops.Plan: Establish the positive coordinate s in the direction the motorcycle is traveling. Since the acceleration is given as a function of time, integrate it once to calculate the velocity and again to calculate the position.
20ò ò EXAMPLE (continued) Solution: 1) Integrate acceleration to determine the velocity.a = dv / dt => dv = a dt =>=> v – vo = -3t2 => v = -3t2 + voò-=tovdtdv)6(2) We can now determine the amount of time required for the motorcycle to stop (v = 0). Use vo = 27 m/s.0 = -3t => t = 3 s3) Now calculate the distance traveled in 3s by integrating the velocity using so = 0:v = ds / dt => ds = v dt =>=> s – so = -t3 + vot=> s – 0 = (3)3 + (27)(3) => s = 54 mò+-=tosdtvds)3(2
21CONCEPT QUIZ t = 2 s t = 7 s 3 m/s 5 m/s 1. A particle moves along a horizontal path with its velocity varying with time as shown. The average acceleration of the particle is _________.A) 0.4 m/s2 B) 0.4 m/s2C) 1.6 m/s2 D) 1.6 m/s2Answers:1. D2. D2. A particle has an initial velocity of 30 ft/s to the left. If it then passes through the same location 5 seconds later with a velocity of 50 ft/s to the right, the average velocity of the particle during the 5 s time interval is _______.A) 10 ft/s B) 40 ft/sC) 16 m/s D) 0 ft/s
22GROUP PROBLEM SOLVINGGiven: Ball A is released from rest at a height of 40 ft at the same time that ball B is thrown upward, 5 ft from the ground. The balls pass one another at a height of 20 ft.Find: The speed at which ball B was thrown upward.Plan: Both balls experience a constant downward acceleration of 32.2 ft/s2. Apply the formulas for constant acceleration, with ac = ft/s2.
23GROUP PROBLEM SOLVING (continued) Solution:1) First consider ball A. With the origin defined at the ground, ball A is released from rest ((vA)o = 0) at a height of 40 ft((sA )o = 40 ft). Calculate the time required for ball A to drop to 20 ft (sA = 20 ft) using a position equation.sA = (sA )o + (vA)ot + (1/2)act220 ft = 40 ft + (0)(t) + (1/2)(-32.2)(t2) => t = s2) Now consider ball B. It is throw upward from a height of 5 ft ((sB)o = 5 ft). It must reach a height of 20 ft (sB = 20 ft) at the same time ball A reaches this height (t = s). Apply the position equation again to ball B using t = 1.115s.sB = (sB)o + (vB)ot + (1/2) ac t220 ft = 5 + (vB)o(1.115) + (1/2)(-32.2)(1.115)2=> (vB)o = 31.4 ft/s
241. A particle has an initial velocity of 3 ft/s to the left at ATTENTION QUIZ1. A particle has an initial velocity of 3 ft/s to the left ats0 = 0 ft. Determine its position when t = 3 s if the acceleration is 2 ft/s2 to the right.A) 0.0 ft B) 6.0 ftC) 18.0 ft D) 9.0 ftAnswers:1. A2. B2. A particle is moving with an initial velocity of v = 12 ft/s and constant acceleration of 3.78 ft/s2 in the same direction as the velocity. Determine the distance the particle has traveled when the velocity reaches 30 ft/s.A) 50 ft B) 100 ftC) 150 ft D) 200 ft