2. Chapter 2 Motion along a straight line

3. 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

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

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

6. Engineering Physics : Lecture 1, Pg 5 Displacement Δx l Displacement: change in position l x 0 = original (initial) location l x = final location l Δx = x - x 0 l Example: x 0 = 1m, x = 4m, l Δx = 4m – 1m = 3m 1m4m

7. Engineering Physics : Lecture 1, Pg 6 Displacement Δx l Displacement: change in position l x 0 = original (initial) location l x = final location l Δx = x - x 0 l Example: x 0 = 4m, x = 1 m, l Δx = 1m – 4m = - 3m 1m4m

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

9. Engineering Physics : Lecture 1, Pg 8 Distance: Scalar Quantity 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.

10. Engineering Physics : Lecture 1, Pg 9 What is the displacement in the situation depicted bellow? l a) 3 m b) 6 m c) -3 m d) 0 m 1m4m

11. Engineering Physics : Lecture 1, Pg 10 What is the distance traveled in the situation depicted bellow? l a) 3 m b) 6 m c) -3 m d) 0 m 1m4m

12. Engineering Physics : Lecture 1, Pg 11 Position, Displacement

13. Engineering Physics : Lecture 1, Pg 12 Motion in 1 dimension l In 1-D, we usually write position as x(t 1 ). Since it’s in 1-D, all we need to indicate direction is + or .  Displacement in a time  t = t 2 - t 1 is  x = x(t 2 ) - x(t 1 ) = x 2 - x 1 t x t1t1 t2t2  x  t x1x1 x2x2 some particle’s trajectory in 1-D

14. Engineering Physics : Lecture 1, Pg 13 1-D kinematics t x t1t1 t2t2  x x1x1 x2x2 trajectory l Velocity v is the “rate of change of position” Average velocity v av in the time  t = t 2 - t 1 is:  t V av = slope of line connecting x 1 and x 2.

15. Engineering Physics : Lecture 1, Pg 14 l Consider limit t 1 t 2 l Instantaneous velocity v is defined as: 1-D kinematics... t x t1t1 t2t2  x x1x1 x2x2  t so v(t 2 ) = slope of line tangent to path at t 2.

16. Engineering Physics : Lecture 1, Pg 15 1-D kinematics... l Acceleration a is the “rate of change of velocity” Average acceleration a av in the time  t = t 2 - t 1 is: l And instantaneous acceleration a is defined as: using

17. Engineering Physics : Lecture 1, Pg 16 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 a v t t t

18. Engineering Physics : Lecture 1, Pg 17 More 1-D kinematics l We saw that v = dx / dt l 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: v(t) t ++...+ = displacement

19. Engineering Physics : Lecture 1, Pg 18 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: 1-D Motion with constant acceleration

20. Engineering Physics : Lecture 1, Pg 19 Recap l So for constant acceleration we find: x a v t t t

21. Engineering Physics : Lecture 1, Pg 20 Example: Displacement vs. Time 8 m 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?

22. Engineering Physics : Lecture 1, Pg 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.

23. Engineering Physics : Lecture 1, Pg 22 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. (c) v = 0, but a  0. y

24. Engineering Physics : Lecture 1, Pg 23 Solution x a v t t t l Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is momentarily zero. l Since the velocity is continually changing there must continually changing there must be some acceleration. 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) l The answer is (c) v = 0, but a  0.

25. Engineering Physics : Lecture 1, Pg 24 Galileo’s Formula l Plugging in for t: l Solving for t:

26. Engineering Physics : Lecture 1, Pg 25 Alternate (Calculus-based) Derivation (chain rule) (a = constant)

27. Engineering Physics : Lecture 1, Pg 26 Recap: l For constant acceleration: l + Galileo and average velocity:

28. Engineering Physics : Lecture 1, Pg 27 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 a b x = 0, t = 0 abab vovo

29. Engineering Physics : Lecture 1, Pg 28 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 a b. At what time t f does the car stop, and how much farther x f does it travel? x = x f, t = t f v = 0 x = 0, t = 0 abab v0v0

30. Engineering Physics : Lecture 1, Pg 29 Problem 1... l Above, we derived: v = v 0 + at l Realize that a = -a b l Also realizing that v = 0 at t = t f : find 0 = v 0 - a b t f or t f = v 0 /a b

31. Engineering Physics : Lecture 1, Pg 30 Problem 1... l To find stopping distance we use: l In this case v = v f = 0, x 0 = 0 and x = x f

32. Engineering Physics : Lecture 1, Pg 31 Problem 1... l So we found that l Suppose that v o = 29 m/s l Suppose also that a b = g = 9.81 m/s 2 ç Find that t f = 3 s and x f = 43 m

33. Engineering Physics : Lecture 1, Pg 32 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).

34. Engineering Physics : Lecture 1, Pg 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!

35. Engineering Physics : Lecture 1, Pg 34 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/s 2, can the train be stopped in time? Know: a =  1.52 m/s 2, v o = 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.

36. Engineering Physics : Lecture 1, Pg 35 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? Example Meeting Example Meeting t o = 0 s, initial time t = final time x o = 44 m v t = 35 m/s, constant v oc = 47 m/s, initial speed of car v c = final speed of car Car’s equation of motion Train’s equation of motion: To meet : x c = x t t = 2.733 s O xoxo x t = xcxc

37. Engineering Physics : Lecture 1, Pg 36 Recap l Scope of this course l Measurement and Units (Chapter 1) çSystems of units çConverting between systems of units çDimensional Analysis l 1-D Kinematics l 1-D Kinematics (Chapter 2) çAverage & instantaneous velocity and acceleration çMotion with constant acceleration l Example car problem