Chapter 2 Motion along a straight line 2

Chapter 2 Motion along a straight line

2. 2 Motion: change in position in relation with an object of reference. The study of motion is called kinematics. Examples: • The Earth orbits around the Sun • A roadway moves with Earth’s rotation

Position and Displacement l Position: coordinate of position – distance to the origin (m) P O x x Engineering Physics : Lecture 1, Pg 3

Example Position 2 1 0 1 x (cm) 2 The position of the ball is The + indicates the direction to the right of the origin. 4 Engineering Physics : Lecture 1, Pg 4

Displacement Δx l l l Displacement: change in position x 0 = original (initial) location x = final location Δx = x - x 0 Example: x 0 = 1 m, x = 4 m, Δx = 4 m – 1 m = 3 m 1 m 4 m Engineering Physics : Lecture 1, Pg 5

Displacement Δx l l l Displacement: change in position x 0 = original (initial) location x = final location Δx = x - x 0 Example: x 0 = 4 m, x = 1 m, Δx = 1 m – 4 m = - 3 m 1 m 4 m Engineering Physics : Lecture 1, Pg 6

Displacement, Distance l l l Distance traveled usually different from displacement. Distance always positive. Previous example: always 3 m. Engineering Physics : Lecture 1, Pg 7

Distance: Scalar Quantity Distance is the path length traveled from one location to another. It will vary depending on the path. Distance is a scalar quantity—it is described only by a magnitude. Engineering Physics : Lecture 1, Pg 8

What is the displacement in the situation depicted bellow? l a) 3 m b) 6 m c) -3 m 1 m d) 0 m 4 m Engineering Physics : Lecture 1, Pg 9

What is the distance traveled in the situation depicted bellow? l a) 3 m b) 6 m c) -3 m 1 m d) 0 m 4 m Engineering Physics : Lecture 1, Pg 1

Position, Displacement Engineering Physics : Lecture 1, Pg 1

Motion in 1 dimension l In 1 -D, we usually write position as x(t 1 ). l Since it’s in 1 -D, all we need to indicate direction is + or . è Displacement in a time t = t 2 - t 1 is x = x(t 2) - x(t 1) = x 2 - x 1 x x some particle’s trajectory in 1 -D x 2 x 1 t 1 t t 2 t Engineering Physics : Lecture 1, Pg 1

1 -D kinematics l l Velocity v is the “rate of change of position” Average velocity vav in the time t = t 2 - t 1 is: x x trajectory x 2 Vav = slope of line connecting x 1 and x 2. x 1 t 2 t t Engineering Physics : Lecture 1, Pg 1

1 -D kinematics. . . l l Consider limit t 1 t 2 Instantaneous velocity v is defined as: x x so v(t 2) = slope of line tangent to path at t 2. x 2 x 1 t 2 t t Engineering Physics : Lecture 1, Pg 1

1 -D kinematics. . . l Acceleration a is the “rate of change of velocity” Average acceleration aav in the time t = t 2 - t 1 is: l And instantaneous acceleration a is defined as: l using Engineering Physics : Lecture 1, Pg 1

Recap l If the position x is known as a function of time, then we can find both velocity v and acceleration a as a function of time! x v a t t t Engineering Physics : Lecture 1, Pg 1

More 1 -D kinematics l We saw that v = dx / dt In “calculus” language we would write dx = v dt, which we can integrate to obtain: l Graphically, this is adding up lots of small rectangles: l v(t) + +. . . + = displacement t Engineering Physics : Lecture 1, Pg 1

1 -D Motion with constant acceleration l High-school calculus: l Also recall that l Since a is constant, we can integrate this using the above rule to find: l Similarly, since we can integrate again to get: Engineering Physics : Lecture 1, Pg 1

Recap l So for constant acceleration we find: x v a t t t Engineering Physics : Lecture 1, Pg 1

Example: Displacement vs. Time b) What is the velocity at 5 s? Unable to determine (no slope) c) What is the position after 10 s? -4 m d) What is the velocity at 10 s? 0 m/s e) What is the velocity during the second part of the trip? 8 m Engineering Physics : Lecture 1, Pg 2

f) What is the velocity during the forth part of the trip? g) What is the velocity at 15 s? h) What is the position after 15 s? 2 m. The displacement changes 1 m every 2 seconds, so the position at 15 s is one meter more than the position at 13 s. Engineering Physics : Lecture 1, Pg 2

Motion in One Dimension l When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path? (a) Both v = 0 and a = 0. (b) v 0, but a = 0. y (c) v = 0, but a 0. Engineering Physics : Lecture 1, Pg 2

Solution l Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is momentarily zero. x l Since the velocity is continually changing there must v be some acceleration. çIn fact the acceleration is caused by gravity (g = 9. 81 m/s 2). ç(more on gravity in a few lectures) a l The answer is (c) v = 0, but a 0. t t t Engineering Physics : Lecture 1, Pg 2

Galileo’s Formula l Solving for t: l Plugging in for t: Engineering Physics : Lecture 1, Pg 2

Alternate (Calculus-based) Derivation (chain rule) (a = constant) Engineering Physics : Lecture 1, Pg 2

Recap: l For constant acceleration: l + Galileo and average velocity: Engineering Physics : Lecture 1, Pg 2

Problem 1 l A car is traveling with an initial velocity v 0. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab vo ab x = 0, t = 0 Engineering Physics : Lecture 1, Pg 2

Problem 1. . . l A car is traveling with an initial velocity v 0. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab. At what time tf does the car stop, and how much farther xf does it travel? v 0 ab x = 0, t = 0 v=0 x = xf , t = t f Engineering Physics : Lecture 1, Pg 2

Problem 1. . . l Above, we derived: v = v 0 + at l Realize that a = -ab l Also realizing that v = 0 at t = tf : find 0 = v 0 - ab tf or tf = v 0 /ab Engineering Physics : Lecture 1, Pg 2

Problem 1. . . l To find stopping distance we use: l In this case v = vf = 0, x 0 = 0 and x = xf Engineering Physics : Lecture 1, Pg 3

Problem 1. . . l l l So we found that Suppose that vo = 29 m/s Suppose also that ab = g = 9. 81 m/s 2 ç Find that tf = 3 s and xf = 43 m Engineering Physics : Lecture 1, Pg 3

Tips: l Read ! çBefore you start work on a problem, read the problem statement thoroughly. Make sure you understand what information is given, what is asked for, and the meaning of all the terms used in stating the problem. l Watch your units ! çAlways check the units of your answer, and carry the units along with your numbers during the calculation. l Understand the limits ! çMany equations we use are special cases of more general laws. Understanding how they are derived will help you recognize their limitations (for example, constant acceleration). Engineering Physics : Lecture 1, Pg 3

Problem 2 It normally takes you 10 min to travel 5 km to school. You leave class 15 min before class. Delays caused by traffic slows you down to 20 km/h for the first 2 km of the trip, will you be late to class? Not late! Engineering Physics : Lecture 1, Pg 3

Example: A train of mass 55, 200 kg is traveling along a straight, level track at 26. 8 m/s. Suddenly the engineer sees a truck stalled on the tracks 184 m ahead. If the maximum possible braking acceleration has magnitude 1. 52 m/s 2, can the train be stopped in time? Know: a = 1. 52 m/s 2, vo = 26. 8 m/s, v = 0 Using the given acceleration, compute the distance traveled by the train before it comes to rest. The train cannot be stopped in time. 34 Engineering Physics : Lecture 1, Pg 3

Example Meeting A train is moving parallel and adjacent to a highway with a constant speed of 35 m/s. Initially a car is 44 m behind the train, traveling in the same direction as the train at 47 m/s, and accelerating at 3 m/s 2. What is the speed of the car just as it passes the train? xo xt = O to = 0 s, initial time t = final time xo = 44 m vt = 35 m/s, constant voc = 47 m/s, initial speed of car vc = final speed of car To meet: Car’s equation of motion xc Train’s equation of motion: xc = xt t = 2. 733 s Engineering Physics : Lecture 1, Pg 3

Recap l l Scope of this course Measurement and Units (Chapter 1) çSystems of units çConverting between systems of units çDimensional Analysis 1 -D Kinematics (Chapter 2) çAverage & instantaneous velocity and acceleration çMotion with constant acceleration Example car problem Engineering Physics : Lecture 1, Pg 3