1 – D Motion

1. 1 – D Motion • We begin our discussion of kinematics (description of motion in mechanics) • Simplest case: motion of a particle in 1 – D • Concept of a particle: idealization of treating a body as a single point (we get close to doing this by pinpointing the “license plate of a car,” etc.) • 1 – D motion: Only two possible directions • We will be working with vector quantities • Characterized by both a magnitude AND a direction • Scalar quantities have a magnitude but NOT a direction • How do we describe motion in physics? • Should be able to answer questions, “Where is it?” along with “Where is it going?” and “How fast is it going?”

2. +x Position vectors (keep track of position of car) Displacement vector (change in position from point 1 to point 2) Position and Displacement • Example: Motion of a car along Sandusky St. (north) Park Ave. (reference point or origin) William St. (point 1) Central Ave. (point 2)

3. CQ1: Consider the figure below that shows three paths between position 1 and position 2. (Path C is a half circle.) Which path would result in the greatest displacement for a particle moving from position 1 to position 2? • Path A • Path B • Path C • All paths would result in the same displacement.

4. Stopwatch measures time t1 Stopwatch measures time t2 (average speed = total distance traveled / time to travel that distance) Velocity +x • Velocity describes speed (magnitude) and direction • Need time as well as position information • Average velocity: Park Ave. (reference point or origin) William St. (point 1) Central Ave. (point 2)

5. Stopwatch measures time t2 Stopwatch measures time t1 Velocity • Average velocity could be positive or negative • Trip from William St. to Central Ave. (previous picture) – assume t1= 0 s and t2= 40 s: • Trip from Central Ave. to William St.: +x Park Ave. (origin) William St. (point 2) Central Ave. (point 1)

6. Trip #1 Trip #2 vaveis the same for both trips! t1 t2 Velocity • Assume t1 = 0 and t2 = 40 s: • Ave. velocity depends only on the total displacement that occurs during time interval Dt= t2 – t1, not on what happens during Dt: x Central Ave. William St. time Park Ave.

7. Velocity • Note that each line represents how position changes with time and is not representative of a path in space • Slope of line between 2 points on this graph gives average velocity between these points • Instantaneous velocity = velocity at any specific instant in time or specific point along path • Mathematically, it is the limit as Dt 0 of the average velocity: • Say we wanted to know velocity at time t1 • Imagine moving t2 closer and closer to t1 • Dx & Dt become very small, but ratio not necessarily small

8. Slope = instant. velocity at timet2 Slope = instant. velocity at timet1 t1 t2 Instantaneous Velocity • Graphically, instantaneous velocity at a point = slope of line tangent to curve at that point x Central Ave. William St. time Park Ave.

9. Velocity • Be careful not to interchange “speed” with “velocity” • “speed” refers to a magnitude only • “velocity” refers to magnitude and direction • For example, I might run completely around a 1–mi. circular track in ¼ hour, with an average speed (round-trip) = dist. traveled / time = 1 mi / 0.25 hr = 4 mph • BUT since • In general: instantaneous speed = but average speed  (I’m right back where I started!)

10. CQ2: The graph below represents a particle moving along a line. What is the total distance traveled by the particle from t = 0to t = 10seconds? • 0 m • 50 m • 100 m • 200 m

11. Slope = average acceleration v2 v1 t1 t2 Acceleration • Acceleration describes changes in velocity • Rate of change of velocity with time • Vector quantity (like velocity and displacement) • “Velocity of the velocity” • Average acceleration defined as: • are the instantaneous velocities at t2 and t1, respectively • To find average acceleration graphically: v t

12. Slope of tangent line = instantaneous acceleration v1 t1 Acceleration • We can also find the acceleration at a specific instant in time, similar to velocity • Instantaneous acceleration: • Graphically: • BE CAREFUL with algebraic sign of acceleration – negative sign does not always mean that body is slowing down v t

13. Acceleration • Example: Car has neg. acceleration if slowing down from pos. velocity, or if it’s speeding up in neg. direction (going in reverse) • Must compare direction of velocity and acceleration: • Be careful not to interchange deceleration with negative acceleration Same sign Oppositesign (for example, going slower in reverse)

14. CQ3: Which animation shown in class gave the correct position vs. time graph? (Assume that the positive x direction is to the right.) • Animation 1 • Animation 2 • Animation 3 • Animation 4 PHYSLET #7.1.1, Prentice Hall (2001)

15. Constant Acceleration v • For special case of constant acceleration, then v vs. t graph becomes a straight line • Note that slope of above line (= acceleration) is the same for any line determined from 2 points on the line or for any tangent line since (for this special case) • We can find the instantaneous acceleration from:  Constant acceleration means constant rate of increase for v t

16. Average value of v: (since v is a linear function) Also, by definition: Setting (2) = (3), and using (1), we get: Constant Acceleration • Now let t1= 0 (can start our clock whenever we want) and t2 =t (some arbitrary time later), and define v(t1= 0)=v0 and v(t2= t) = v •  • Graphical interpretation: (takes form of equation of a line:y = b + mx) v v at v0 v0 t

17. Constant Acceleration • We can use equations (1) and (4) together to obtain an equation independent of t and another equation independent of a using substitution, with the result: (independent of t) (independent of a)

18. Constant Acceleration • Graphs of a, v, and x versus time: Motion Graphs Interactive

19. CQ4: The graph below represents a particle moving along a line. When t = 0, the displacement of the particle is 0. All of the following statements are true about the particle EXCEPT: • The particle has a total displacement of 100 m. • The particle moves with constant acceleration from 0 to 5 s. • The particle moves with constant velocity between 5 and 10s. • The particle is moving backwards between 10 and 15 s.

20. Interactive Example Problem:Seat Belts Save Lives! Animation and solution details given in class. ActivPhysics Problem #1.8, Pearson/Addison Wesley (1995–2007)

21. Example Problem #2.39 • A car starts from rest and travels for 5.0 s with a uniform acceleration of +1.5 m/s2. The driver then applies the brakes, causing a uniform acceleration of –2.0 m/s2. If the brakes are applied for 3.0 s, • how fast is the car going at the end of the braking period, and • how far has the car gone? • Solution (details given in class): • 1.5 m/s • 32 m

22. Example Problem #2.43 • A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 12 m/s, skates by with the puck. After 3.0 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 4.0 m/s2, • how long does it take him to catch his opponent, and • how far has he traveled in that time? (Assume that the player with the puck remains in motion at constant speed.) • Solution (details given in class): • 8.2 s • 134 m

23. Free Fall • Particular case of motion with constant acceleration: “free-fall” due to gravity • Objects falling to the ground • Galileo (16th Century) showed that bodies fall with constant downward acceleration, independent of mass (neglecting air resistance and buoyancy effects) • Near the Earth’s surface, this acceleration has a constant value ofg 9.8 m/s2 • We can use all of the constant-acceleration equations for free fall • Will use “y” instead of “x” when referring to vertical positions:

24. Free Fall +y(up) • Sign of depends on choice of +y – axis direction: • Choice of axis not important – as long as you are consistent! • always points toward the ground, however  then = 0 0  then = +y(down)

25. CQ5: A ball is thrown straight up in the air. For which situation are both the instantaneous velocity and the acceleration zero? • on the way up • at the top of the flight path • on the way down • halfway up and halfway down • none of these

26. +y 443 m • y = y0 + v0t – ½ gt2 • – 443 m = 0 + 0 – ½ gt2 • 886 m / g = t2 t2 = 90.4 s2 • t =  9.5 s t = +9.5 s (neg. value has no physical meaning) Free Fall Example Problem How long would it take for Tom Petty to go “Free-Fallin’” from the top of the Sears Tower in Chicago? (Of course there is a safety net at the bottom.) How fast will he be moving just before he hits the net? • v = v0 – gt = 0 – gt = – (9.8 m/s2)(9.5 s) = – 93.1 m/s (negative sign means downward direction)

27. CQ6: A pebble is dropped from rest from the top of a tall cliff and falls 4.9 m after 1.0 s has elapsed. How much farther does it drop in the next 2.0 seconds? • 9.8 m • 19.6 m • 39 m • 44 m • 27 m

28. Final Comments About 1–D Motion v Free fall only vtis called the “terminal velocity” vt • In reality, Tom Petty’s velocity would “tail off” and start reaching a maximum: • Other notes about 1–D velocity and acceleration: • In general, both velocity and acceleration are not constant in time • Can use methods of calculus to generalize equations • General formulas reduce to constant acceleration formulas when a(t)= constant =a • Remember that formula d = vt only holds when v = constant! Including air resistance effects t