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Chapter 2

Chapter 2. Motion along a straight line. 2.2 Motion. 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.

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Chapter 2

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  1. Chapter 2 Motion along a straight line

  2. 2.2 Motion 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

  3. Position and Displacement Position: coordinate of position – distance to the origin (m) P O x x

  4. Example Position x (cm) 1 2 0 1 2 The position of the ball is The + indicates the direction to the right of the origin. 4

  5. Displacement Δx Displacement: change in position x0 = original (initial) location x = final location Δx = x - x0 Example: x0 = 1m, x = 4m, Δx = 4m – 1m = 3m 1m 4m

  6. Displacement Δx Displacement: change in position x0 = original (initial) location x = final location Δx = x - x0 Example: x0 = 4m, x = 1 m, Δx = 1m – 4m = - 3m 1m 4m

  7. Displacement, Distance Distance traveled usually different from displacement. Distance always positive. Previous example: always 3 m.

  8. 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.

  9. What is the displacement in the situation depicted bellow? a) 3 m b) 6 m c) -3 m d) 0 m 1m 4m

  10. What is the distance traveled in the situation depicted bellow? a) 3 m b) 6 m c) -3 m d) 0 m 1m 4m

  11. Position, Displacement

  12. Motion in 1 dimension • In 1-D, we usually write position as x(t1 ). • Since it’s in 1-D, all we need to indicate direction is + or . • Displacement in a time t = t2 - t1isx = x(t2) - x(t1) = x2 - x1 x some particle’s trajectoryin 1-D x2 x x1 t1 t2 t t

  13. 1-D kinematics • Velocity v is the “rate of change of position” • Average velocity vav in the time t = t2 - t1is: x trajectory x2 x Vav = slope of line connecting x1 and x2. x1 t1 t2 t t

  14. 1-D kinematics... • Considerlimit t1 t2 • Instantaneous velocity v is definedas: x sov(t2) = slope of line tangent to path at t2. x2 x x1 t1 t2 t t

  15. 1-D kinematics... • Acceleration a is the “rate of change of velocity” • Average acceleration aavin the time t = t2 - t1is: • Andinstantaneous acceleration a is definedas: using

  16. Recap • If the position x is known as a function of time, then we can find both velocity vand acceleration a as a function of time! x t v t a t

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

  18. 1-D Motion with constant acceleration • High-school calculus: • Also recall that • Since a is constant, we can integrate this using the above rule to find: • Similarly, since we can integrate again to get:

  19. Recap • So for constant acceleration we find: x t v t a t

  20. 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

  21. 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.

  22. Motion in One Dimension • 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)Bothv = 0anda = 0. (b)v  0, but a = 0. (c) v = 0, but a  0. y

  23. Solution • Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is momentarily zero. • Since the velocity is continually changing there must be some acceleration. • In fact the acceleration is caused by gravity (g = 9.81 m/s2). • (more on gravity in a few lectures) • The answer is (c) v = 0, but a  0. x t v t a t

  24. Solving for t: Galileo’s Formula • Plugging in for t:

  25. (chain rule) (a = constant) Alternate (Calculus-based) Derivation

  26. Recap: • For constant acceleration: • + Galileo and average velocity:

  27. vo ab x = 0, t = 0 Problem 1 • A car is traveling with an initial velocity v0. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab

  28. Problem 1... • A car is traveling with an initial velocity v0. 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? v0 ab x = 0, t = 0 v = 0 x = xf , t = tf

  29. Problem 1... • Above, we derived: v = v0 + at • Realize thata = -ab • Also realizing that v = 0at t = tf : find 0 = v0 - ab tfor tf = v0 /ab

  30. Problem 1... • To find stopping distance we use: • In this case v = vf = 0, x0 = 0 and x = xf

  31. Problem 1... • So we found that • Suppose that vo = 29 m/s • Suppose also thatab = g = 9.81 m/s2 • Find that tf = 3 s and xf = 43 m

  32. Tips: • Read ! • Before 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. • Watch your units ! • Always check the units of your answer, and carry the units along with your numbers during the calculation. • Understand the limits ! • Many 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).

  33. 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!

  34. 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/s2, can the train be stopped in time? Know: a = 1.52 m/s2, 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

  35. 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/s2. What is the speed of the car just as it passes the train? xo xt = O xc Car’s equation of motion 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 Train’s equation of motion: To meet: xc = xt t = 2.733 s

  36. Recap • Scope of this course • Measurement and Units (Chapter 1) • Systems of units • Converting between systems of units • Dimensional Analysis • 1-D Kinematics (Chapter 2) • Average & instantaneous velocity and acceleration • Motion with constant acceleration • Example car problem

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