Day 2 UNIT 1 Motion Graphs x t Lyzinski Physics.

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Days UNIT 1 Motion Graphs x t Lyzinski Physics.
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Presentation transcript:

Day 2 UNIT 1 Motion Graphs x t Lyzinski Physics

Day #2 * Position * Displacement * Average Velocity * Vectors * x-t graphs

Definition Position (x) – the location of an object with respect to a specified reference point. *We choose this reference point to be the origin of a coordinate system. -3 -2 -1 0 1 A km 6 7 8 9 10 The position of particle “A” is either x = -3 or x = 6, depending on which reference point (or origin) you use.

These are all VECTORS. What’s a vector? Definition Displacement (Dx) – the change in an object’s position during a time interval. Dx = x2 – x1 or Dx = xf – xi *Displacement must have both a magnitude (size) and a direction (right, left, up, down, north, south, etc). These are all VECTORS. What’s a vector?

1m, 1m [right] 4m, 4m [left] 8m, 0m 5m, 3m [left] -3 -2 -1 0 1 B C A -3 -2 -1 0 1 A meters Using the number line above, find the distance travelled and the displacement in moving from 1m, 1m [right] - A to B - C to A - A to C and then back to A - C to B, passing through A Dx = 1 – (1m) = 0m 4m, 4m [left] 8m, 0m 5m, 3m [left] Dx = (-2) – (1m) = -3m OR 3m [left]

(or v=Dx/t) Definition Average Velocity ( v ) – the displacement of an object divided by the elapsed time. v = Dx/Dt (or v=Dx/t)

Definition position velocity Vector – a quantity that has both magnitude AND a direction … oh yeh! * YES, vectors can have units. ** What vectors have we learned about thus far? ____________ ________________ ___________ position displacement velocity

Scalars vs. Vectors Displacement: has magnitude & direction (example: 15 cm east) Distance: has a magnitude only (example: 6 ft) 1 2 A B Displacement is NEVER greater than distance traveled!

Scalars vs. Vectors (continued) Velocity: has magnitude & direction (example: 15 mi/h North) Speed: has a magnitude only (example: 30 km/h) 1 2 Total time for the trip from 1 to 2: 2 hr 25 km 16o 24 km 7 km Speed = d/t = 15.5 km/h Velocity = Dx/t = 12.5 km/h If an object STARTS & STOPS at the same point, the velocity is ZERO! (since the displacement is zero)

x-t graphs t (sec) x (m) t1 t2 t3 x2 x1 x3 B C D A Constant speed (Constant + velocity, or constant velocity in the + direction) Slow down, speed up, slow down, speed up 2 moments where the object is “at rest” (for a moment)

How to get the position (x) at a certain time (t) off an x-t graph x (m) 10 20 30 40 50 t (s) 30 20 10 Example: What is the position at t = 30 seconds? 24m Go over to t = 30. Find the pt on the curve. Find the x value for this time.

How to calculate the displacement between two times on an x-t graph x (m) 10 20 30 40 50 t (s) 30 20 10 Example: What is the displacement from t = 10 to t = 40? 17 m Find x1 Find x2 10 m Use D x = x2 - x1 = + 7 m

How to find the distance traveled between two times on an x-t graph. x (m) 10 20 30 40 50 t (s) 30 20 10 Example: What is the distance traveled from t = 10 to t = 40? 17 m 10 m Find the distance traveled in the + direction. Find the distance traveled in the - direction. Add them together. (27 m)

Understand the difference between velocity and speed on an x-t graph. x (m) 10 20 30 40 50 t (s) 30 20 10 Example: What is the average speed from t = 10 to t = 40 seconds? 17 m 10 m dist10-40 = 27 m (previous slide) Avg. Speed = dist/ Dt = 27m / 30 sec = 0.9 m/s

Understand the difference between velocity and speed on an x-t graph. x (m) 10 20 30 40 50 t (s) 30 20 10 Example: What is the average velocity from t = 10 to t = 40 seconds? Dx10-40 = + 7 m (previous slide) Avg. Velocity = slope = Dx/ Dt = + 7 / 30 sec = + 0.23 m/s

Will avg. velocity EVER be greater than avg. speed? NO!!! Will avg. velocity EVER be equal to avg. speed? YES!!! When the path travelled was one-way, in a straight line.

Negative Average Velocity? x (m) 10 20 30 40 50 t (s) 30 20 10 Example: What is the average velocity from t = 20 to t = 40 seconds? Avg. vel. = slope = rise/run = -7 m / 20 = -.35 m/s Since the objects displacement is in the NEGATIVE direction, so is its average velocity.

1 Open to in your GREEN packet -10 m 2) 3) 4) avg velocity = slope = -15m / 6sec = -2.5 m/s s = |v| = 2.5 m/s At rest at t = 0 and t = 12 sec

Speeding up, const negative vel, slowing down, speeding up, 5) 6) Speeding up, const negative vel, slowing down, speeding up, const positive velocity(slow), speeding up, constant positive velocity (fast) Dx = x2 – x1 = (-10m) – (10m) = -20m (approximately)

HOMEWORK Check out your Unit 1 Schedule … Day #2 Again, we will “try” to follow it night by night.