Days 1 - 2 UNIT 1 Motion Graphs x t Lyzinski Physics

The purpose of this chapter is to learn the 1st step of Mechanics (the study of motion) which is KINEMATICS (the study of motion with no regards to what is causing the motion). The study of what is “causing” the motion is known as dynamics, and we will study this in a later chapter.

PHYSICS - Mechanics - Electricity - Magnetism - Optics - Waves KINEMATICS - Mechanics A “description” of motion - Electricity DYNAMICS - Magnetism A study of what “causes” motion - Optics - Waves

Day #1 * Distance * Speed * Scalars * d-t graphs

Definition Distance (d) – the length of the path followed by an object * If an object’s path is straight, the distance is the length of the straight line between start and finish. ** If an object’s path is NOT straight, the distance is the length of the path if you were to “straighten it out” and measure it the way you would measure the length of a curved shoelace. start finish start finish

B C -3 -2 -1 0 1 A meters Using the number line above, what would be the distance travelled if an object travelled from ….. 1m - A to B - A to C - A to C and then back to A - C to B, passing through A 4m 4m + 4m = 8m 4m + 1m = 5m

4 yd B 12 yd 5 yd C A Sally and Timmy are at point A. Sally walks directly to point C (taking the shortest path). Timmy also takes a “shortest path”, but has to stop at point B for lunch first. How much further has Timmy walked when he arrives? 4 yd

Definition Average Speed (s) – the distance travelled during a time interval divided by the elapsed time. s = d/Dt (or s=d/t) Since Dt = t2 – t1, if t1 = 0, then Dt = t2 – 0 = t2 = t

1 h, 10 min d = 3mi + 4mi = 7mi s = 6 mi/h B C -3 -2 -1 0 1 A miles Larry walks from point B to point C, and then goes directly to point A. If he walks at an average speed of 6 mph, how long does the trip take him? d = 3mi + 4mi = 7mi s = 6 mi/h s = d/t  t = d/s = (7mi)/(6mi/h)=1.17h 1 h, 10 min Use appropriate units

B C -3 -2 -1 0 1 A km Larry runs from point A to point B in 5 minutes and then proceeds to jog directly to point C, taking his time in 30 additional minutes. Find… Larry’s average speed during the first portion of the trip. The average speed during the second portion of the trip. Larry’s average speed for the entire trip. s = d/t = (1km)/(5min) = 0.2 km/min = 12 km/h s = d/t = (3km)/(30min) = 0.1 km/min = 6 km/h s = d/t = (4km)/(35min) = 0.114 km/min = 6.86 km/h

Definition distance speed time Scalar – a quantity that has a magnitude only, no direction. * YES, scalars can have units. ** What scalars have we learned about thus far? ___________ ____________ ___________ distance speed time I thought time could march backward?

d-t graphs E F D C B A Constant speed Speeding UP Constant Speed (faster!) Slowing Down At rest t (sec) d (m) B C E A D F 10 15 20 27 36 120 100 50 30

sAB = rise/run = (30-0m) / (10-0s) = 3 m/s t (sec) d (m) B C E A D F 10 15 20 27 36 120 100 50 30 SLOPE Speed on a d-t graph can be found by taking the _______________. sAB = rise/run = (30-0m) / (10-0s) = 3 m/s sCD = rise/run = (100-50m) / (20-15s) = 10 m/s

Open to in your Unit 1 packet 1 d-t graphs CANNOT have sharp points NOTHING CAN STOP INSTANTANEOUSLY!! 520 – 170yd = 350 yd (approximately)

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 4m, 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 4m, 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)

Find Sam’s avg. speed and avg. velocity for the entire trip. B C Sam runs the 400m dash. He starts and finishes at point A, travelling one complete circuit around the track. Each section of the track is 100m long. His average speed during each interval are as follows. AB: 7 m/s BC: 8 m/s CD: 6 m/s DA: 7.5 m/s Find Sam’s avg. speed and avg. velocity for the entire trip. s = d/t  t = d/s = 100m/7sec = 14.286 sec 100m/8sec = 12.5 sec 100m/6sec = 16.667 sec 100m/7.5sec = 13.333 sec s = d/t = (400m)/(56.786s) = 7.04 m/sec Avg Velocity = 0 since Dx = 0 for the entire trip.

s = d/t  t = d/s = 200m/(14.286+12.5s) = 7.47 m/s A D B C Sam runs the 400m dash. He starts and finishes at point A, travelling one complete circuit around the track. Each section of the track is 100m long. His average speeds during each interval are as follows. AB: 7 m/s, 14.286 sec BC: 8 m/s, 12.5 sec CD: 6 m/s, 16.667 sec DA: 7.5 m/s, 13.333 sec 100 104.94 31.831 m Find Sam’s average speed and average velocity for the 1st half of the race. s = d/t  t = d/s = 200m/(14.286+12.5s) = 7.47 m/s v = Dx/t = (104.94m )/(14.286+12.5s) = 3.92 m/sec

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: 1 hr 25 km 16o 24 km 7 km Don’t worry about this notation for this test  Speed = 31 km/h Velocity = 25 km/h at 16o NE 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.

Open to in your Unit 1 packet 1 -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)