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Physics 231

Physics 231. Topic 2: One dimensional motion. Alex Brown August 29, 2014. What’s up? (Wednesday Sept 3 rd ) 1) Pretest due Wednesday Sept 3rd 10 pm (follow email instructions). 2) Homework set 00 due Sunday Sept 7 th 10 pm 3) Homework set 01 due Tuesday Sept 9 th 10 pm

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Physics 231

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  1. Physics 231 Topic 2: One dimensional motion Alex Brown August 29, 2014

  2. What’s up? (Wednesday Sept 3rd) 1) Pretest due Wednesday Sept 3rd 10 pm (follow email instructions). 2) Homework set 00 due Sunday Sept 7th 10 pm 3) Homework set 01 due Tuesday Sept 9th 10 pm 4) LRC (1248 BPS) (starts this Friday) Friday 9-12:30 Monday 9 am to 8:30 pm and Tuesday 9 am to 8:30 pm 5) Will start clicker questions this Friday

  3. What’s up? (Friday Sept 5) 1) Homework set 00 due Sunday Sept 7th 10 pm 2) Homework set 01 due Tuesday Sept 9th 10 pm 3) LRC (1248 BPS) (starts this Friday) Friday 9-12:30 Monday 9 am to 8:30 pm and Tuesday 9 am to 8:30 pm 5) clicker questions today

  4. Key Concepts: 1D Motion • Position and coordinate systems • Displacement vs. Distance • Velocity • Distinguish from speed • Average vs instantaneous velocity • Acceleration • Relationship to velocity • Impact on position in 1D motion • Constant velocity and constant acceleration equations • How to set up & solve problems • Understanding graphical representations of distance, velocity, acceleration • Covers chapter 2 in Rex & Wolfson

  5. Galileo (1564 – 1642) What mathematical forces govern accelerated motion? A wooden and stone sphere are dropped from the tower of pisa. Which one reaches the earth first? Answer: they hit the ground at the same time!

  6. The Inclined Plane Experiment Distance traveled goes with the square of time: x(t)  t2

  7. Coordinate Systems We use coordinate systems as a graphical representation of the relationships of measurements • Distance vs time • Espressos sold per shift • Cell phone minutes used per day

  8. θ x Point Particle Treatment The skier can be represented by a simple, point-like object

  9. Position & Displacement

  10. Position & Displacement

  11. Position & Displacement

  12. Position & Displacement Net displacement is a vector and has direction (sign) Net displacement for the entire trip is 60 + 200 - 200 = 60 m Total distance is the sum of the magnitude for each part 60 + 200 + 200 = 460 Displacement is not equal to Distance

  13. Position vs. Time Equally-spaced photographs of a runner reveal his position as a function of time

  14. Position vs. Time Equally-spaced photographs of a runner reveal his position as a function of time

  15. Position vs Time Graph These measurements can be translated to a 2D “Position vs Time” coordinate system.

  16. Average Velocity Displacement Average velocity = Time interval x = 13.6m x = 20.4m t = 2.0s t = 2.0s

  17. Average Velocity Displacement Average velocity= Time interval

  18. X=-30 t=0 t=1 X=0 X=30 X=45 Vectors vs Scalars Average velocity: vector = +30 Average speed: scalar

  19. Average speed: 120 km / (1 + 1.5) hrs = 48 km/h Average velocity: 0 km / (1 + 1.5) hrs = 0 km/h Speed vs Velocity A driver commutes from home in the suburbs to work in the city (each way = 60 km). The morning trip takes 60 minutes (1 hour). The drive home takes 90 minutes (1.5 hours). What was his average speed and average velocity over The whole trip? • Average speed: 60 km/h Average velocity: 48 km/h • Average speed: 48 km/h Average velocity: 0 km/h • Average speed: 60 km/h Average velocity: 0 km/h • Average speed: 0 km/h Average velocity: 60 km/h • No way to tell.

  20. Instantaneous Velocity “But officer, my average speed was only 25 miles per hour...” Sometimes we want to know the speed at one particular point in time.

  21. Instantaneous Velocity The velocity at a single point in time is given by the slope of the line tangent to the x-t curve at that time. Conceptualize this condition as a distance measurement in a very tiny time interval

  22. Instantaneous Velocity Example Draw a tangent line, then put a triangle with sides x and t underneath it (ratio does not depend on the size) x t What is the velocity at point C? x v = x/t = (105-85)/(3-1) = 20/2 = 10 m/s What is the velocity at point E? t v = x/t = (35-75)/(4.5-3.5) = - 40/1 = - 40 m/s Be careful with the sign: velocity is a vector and can be negative!

  23. v …means a change in velocity: Acceleration t Acceleration x A change in slope in the x-t graph… t

  24. Acceleration vf tf Average acceleration: average change in velocity between B and F v vi ti t

  25. Instantaneous Acceleration acceleration at one point in time: Slope of the tangent to the v-t curve at that point in time at D v t

  26. Question t = 0 v = 20 v t = 1 v = 5 t What is the average acceleration between t=0 and t=1 A) 15 B) -15 C) 0 D) infinity

  27. Relationship of Acceleration, Velocity and Displacement

  28. Acceleration of a Car 200m A (m/s2) V (m/s) D (m) Time (s) Time (s) Time (s) 60 mph = 27 m/s

  29. Clicker Question Which v-t diagram matches the x-t diagram on the right?

  30. Question Which v-t diagram matches which a-t diagram?

  31. Overview up to this point VECTORS • Displacement • Average Velocity • Instantaneous velocity • Average acceleration • Instantaneous acceleration SCALARS • Distance • Average speed • Instantaneous speed MOTION DIAGRAMS Next – equations of motion in 1d

  32. Constant Acceleration v(t) v0 velocity v (m/s) Velocity at time t equals… t0 t time (s) Velocity at t=t0=0 Plus the gain in velocity per second multiplied by the time elapsed (every second, the velocity increases with a m/s)

  33. Constant Acceleration Start position plus average velocity multiplied by time Substitute Substitute

  34. Constant Acceleration

  35. (A) (B) (C) (D) (E) You can get C, D and E from A and B

  36. Constant position Constant velocity Constant acceleration x(t)=x0 x(t)=x0+v0t x(t)=x0+v0t+½at2 x(m) x(m) x(m) t t t v m/s v m/s v m/s v(t)=0 v(t)=v0 v(t)=v0+at t t t a m/s2 a m/s2 a m/s2 a(t)=0 a(t)=0 a(t)=a0 t t t

  37. Constant position Constant velocity Constant acceleration x(t)=x0 x(t)=x0+v0t x(t)=x0+v0t+½at2 x(m) x(m) x(m) t t t v m/s v m/s v m/s v(t)=0 v(t)=v0 v(t)=v0+at t t t a m/s2 a m/s2 a m/s2 a(t)=0 a(t)=0 a(t)=a0 t t t

  38. Constant position Constant velocity Constant acceleration x(t)=x0 x(t)=x0+v0t x(t)=x0+v0t+½at2 x(m) x(m) x(m) t t t v m/s v m/s v m/s v(t)=0 v(t)=v0 v(t)=v0+at t t t a m/s2 a m/s2 a m/s2 a(t)=0 a(t)=0 a(t)=a0 t t t

  39. Galileo’s Free Fall Experiment

  40. Gravitational Free Fall 1 kg 5 kg A B 100 m a is the acceleration felt due to gravitation (it is called ‘g’ and its value is 9.81 m/s2) g does not depend on the mass of falling object

  41. Question • A person throws a ball straight up in the air. • Which of the following is true? • The velocity at the highest point is 0. • The acceleration at the highest point is 0. • Both velocity and acceleration at the highest point are 0. • d) Neither velocity nor acceleration are 0 at the highest point.

  42. Ball Toss Problem Dave throws a ball straight up with a velocity of 5.0 m/s. 1) How high does it go relative to the height at which it was released? 2) After how much time does it reach that height? x(t) = x0 +v0t + ½at2 v(t) = v0 + at x(t) = 0 + 5t + ½(-9.8)t2 it decelerates v(ttop) = 0 = 5 - 9.8ttopat it highest point, the velocity is zero ttop = 5/9.8 = 0.51 s time to reach highest point xtop = x(t=0.51s) = 50.51 - ½9.8(0.51) 2 = 1.3 m

  43. V (m/s) t (s) Area indicated by The area under the v-t curve is equal to the displacement (change of position) of the object If the area is negative then the change of position is negative.

  44. a(m/s2) 0 0 t (s) Area indicated by The area under the a-t curve is equal to the change of velocity. If the area is negative then the change of velocity is negative (its slowing down).

  45. V vs T Diagram Example The figure shows velocity versus time. a) What is the displacement after 4 s? b) What is the average velocity? • The distance covered is the area under the v-t diagram. • x = ½x2x4 + 1x4 + (2+½x1x2) = 11 m b) Average velocity = x/t = 11m/4s = 2.75 m/s

  46. A vs T Diagram Example a (m/s2) • Given this a-t diagram, calculate: • Velocity after 4 s • Distance traveled in 4 s • The object was originally at rest. 3 2 1 t (s) 0 1 2 3 4 • v(t) = v0+at (area) • After 2 seconds: v(2) = 0 + 2x1 = 2 m/s • After 4 seconds: v(4) = v(2) + 2x2 = 2 + 4 = 6 m/s b) for x(t) use area under v-t curve After 2 seconds: x(2) = 0 + ½x2x2= 2 m After 4 seconds: x(4) = x(2) = 2 + ½(2x4) + 2x2 = 10 m

  47. Galileo’s Free Fall Experiment Galileo throws a canon ball from the tower of Pisa. Ignoring friction what is the distance covered between t=1 and t=3 seconds? The initial velocity was 0 m/s. x(t) = xo + vot + ½at2 xo = h (height of tower) vo = 0 (started at rest) a = -g (9.81 m/s2 in the –x direction) = xo + vot + ½(-g)t2 = h + - ½gt2 x1 = x(1) - h= - ½9.812 = -4.9 m x3 = x(3) – h = - ½9.832 = -44.1 m Distance traveled: x = | [x(3)-h] - [x(1)-h] | = |-44.1 + 4.9| = 39.2 m

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