Physics Unit 1 Motion Graphs

5 Main Branches of Physics KINEMATICS - Mechanics A “description” of motion - Electricity DYNAMICS - Magnetism A study of what “causes” motion - Optics - Waves

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.

Scalar vs. Vector Scalar – a quantity that has a magnitude only, no direction. Ex: time (5 hours) age (17 years) temperature (20˚C) distance (20 miles) * YES, scalars have units. Vector – a quantity that has both magnitude AND a direction. Ex: displacement (10 m [S]) force (5 N [E])

Distance vs. Displacement Distance (d) – the length of the path followed by an object (scalar) * 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

Displacement – the change in an object’s position during a time interval. (vector) OR the length of a straight line from start to finish. Displacement ≠ distance. However, sometimes the magnitude will be the same. It doesn’t matter what path you take from your house to school, displacement will never change but distance will. *Displacement must have both a magnitude (size) and a direction (right, left, up, down, north, south, etc).

1m, 1m [right]* *notice a direction is given for displacement B C -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]* *notice a direction is given for displacement - 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 no direction needed here 4m, 3m [left]* + or – can be used instead of R and L Dx = (-2) – (1m) = -3m OR 3m [left]

(or s=dist/t) Speed vs. Velocity Average Speed (s) – the distance travelled during a time interval divided by the elapsed time. (scalar) s = dist/Dt (or s=dist/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

v = displace/Dt (or v=d/t) *This line means it’s a vector Average Velocity ( v ) – the displacement of an object divided by the elapsed time. (vector) Avg. velocity is a change in position over a change in time. Since displacement ≠ distance, velocity ≠ speed. v = displace/Dt (or v=d/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. He ended in the exact location he started!!

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 Use Pythagorean theorem to determine the displacement from A to C

Scalar vs. Vector What scalars have we learned about thus far? distance speed time What vectors have we learned about thus far? 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 Don’t worry about this notation for this test just give the general direction  24 km 7 km 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)

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 Graphing distance vs. time SLOPE Speed on a d-t graph can be found by taking the _______________. This gives us the change in distance of an object over a change in time. sAB = rise/run = (30-0m) / (10-0s) = 3 m/s sCD = rise/run = (100-50m) / (20-15s) = 10 m/s

d-t graphs (distance-time) Constant speed Speeding UP –notice how more distance is covered each second Constant Speed (but faster than AB) Slowing Down–less dist. covered each second At rest –no distance covered, but time goes by t (sec) d (m) B C E A D F 10 15 20 27 36 120 100 50 30

Practice Graph 520 – 170yd = 350 yd (approximately) FYI -d-t graphs CANNOT have sharp points. That would mean you came to a stop instantaneously without slowing down first. minutes 520 – 170yd = 350 yd (approximately)

x-t graphs (position – time graphs. Like d-t graphs) (displacement) x-t graphs (position – time graphs. Like d-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) imagine slowing your car to a stop, then going in reverse

How to get the displacement/position (d) at a certain time (t) off an d-t graph. Sometimes I’ll refer to this as x-t d(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 xi Find xf 10 m Use D x = xf - xi = + 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 to the highest point. Find the distance traveled in the – direction from the highest point. 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 Notice the + sign. It indicates direction.

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.

Graph Example -10 m avg velocity = slope = -15m / 6sec = -2.5 m/s 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)

Definition Instantaneous Velocity (v) – the velocity of an object at a precise moment in time.

Just what is meant by “instantaneous” velocity? Dt Dt Dt Dt Finally, “in the limit” that the time interval is infinitely small (or approximately zero), we find the velocity at a single moment in time.  Hence the term “instantaneous velocity” Dt To find the average velocity between two points in time, we find the slope of the line connecting these two points, thus finding the change in position (rise) over the change in time (run). As the two points move closer together, we find the average velocity for a smaller time interval. As the two points move EVEN CLOSER together, we find the average velocity for an EVEN SMALLER time interval.

To find instantaneous velocities, we still use the concept of slope To find instantaneous velocities, we still use the concept of slope. But we find the slope OF THE TANGENT TO THE CURVE at the time in question Definition Tangent to a Curve – a line that intersects a given curve at exactly one point.

Good Tangents  Bad Tangents 

How to find the instantaneous velocity of a specific time interval from an x-t graph … x(m) 10 20 30 40 50 t (s) 30 20 10 Example: What is the instantaneous velocity at t = 20 seconds? (24, 30) (15, 15) Draw the tangent to the curve at the point in question. Then, find the slope of the tangent. Slope = rise/run = 15 m / 9 s = 1.7 m/s (approx) YOU MUST SPECIFY WHICH POINTS YOU USED WHEN FINDING THE SLOPE!!!!

How to find the instantaneous velocity of a specific time interval from an x-t graph … x(m) 10 20 30 40 50 t (s) 30 20 10 Example: What is the instantaneous velocity at t = 5? If the pt lies on a segment, find the slope of the segment. Slope = 5 m / 10 s = 0.5 m/s YOU MUST SPECIFY WHICH POINTS YOU USED WHEN FINDING THE SLOPE!!!! (0,5) (10,10)

How to find the instantaneous velocity of a specific time interval from an x-t graph … x(m) 10 20 30 40 50 t (s) 30 20 10 Example: What is the instantaneous velocity at t = 25 seconds? Draw the tangent to the curve at the point in question. Then, find the slope of the tangent. Slope = 0 (object at rest)

x-t graphs 2 1 3 x (m) x2 x1 x3 t (sec) t1 t2 t3 Slope of line segment Slope of line segment Slope of line segment

Open to in your Unit 1 packet 1 (13.5,-20) (0,6) (33,2) (11,-20) Tangent to the curve has a slope of +22m / 22sec = 1 m/s Tangent to the curve has a slope of -26m / 13.5s = -1.93 m/s THEREFORE, v = -1.93 m/s and s = 1.93 m/s (approximately)

Definition a = vf-vi OR a = Dv/t Average Acceleration ( a ) – the change in an object’s velocity in a given time interval……..IN OTHER WORDS, the rate of change of an object’s velocity. When you stop a car, you actually push the break petal a few seconds before coming to a complete stop. The velocity gradually slows. a = vf-vi OR a = Dv/t tf-ti

Find the acceleration of each object An object is moving at 40 m/s when it slows down to 20 m/s over a 10 second interval. An object is moving at -40 m/s, and 5 seconds earlier it was moving at -10 m/s. An object travelling at -10 in/min is moving at +20 in/min 2 minutes later. 4) An object moving at -30 mph is moving at -20 mph 10 hours later. Slows down Negative accel a = Dv/Dt = (20 – 40m/s) / 10sec = -2 m/s2 Speeds up Negative accel a = Dv/Dt = [-40 – (-10m/s)] / 5sec = -6 m/s2 a = Dv/Dt = [20 – (-10 in/min)] / 2min = +15 in/min2 Speeds up + Accel Slows down + Accel a = Dv/Dt = [-20 – (-30 mi/h)] / 10hr = + 1mi/h2

How can something have a negative acceleration when traveling in a positive direction? When a train, traveling in a positive direction (right) slows as it approaches the next station, velocity can still be +, but acceleration will be negative because initial velocity is larger than final velocity. ∆v is negative. But be careful… Negative acceleration doesn’t always mean deceleration. Think of a train moving in a negative direction (in reverse, or just to the left). Acceleration would be negative when the train gained speed and positive when the train lost speed to enter a station.

Did the last line confuse you Did the last line confuse you? How can something slow down and have a positive acceleration? This example may help An object moving at -30 mph is moving at -20 mph 10 hours later. Its speed (30mph vs. 20 mph) clearly decreases. * remember, speed is |v| As time marches on, the velocity become MORE positive (b/c -20mph is more positive than - 30mph)  THEREFORE, Dv is +

What do the “units” of acceleration mean? m/s2 m/s/s m/s per second t (sec) v (m/s) 1 3 2 6 9 4 12 3 m/s2 means that your velocity increased by 3 m/s every second. t (min) v (km/min) 0.3 1 0.2 2 0.1 3 4 -0.1 -0.1 km/min2 means that your velocity decreased by 0.1 km/min every minute that you were moving.

Really just tangents to the curve at a point. The Key Equations Displacement: ∆d = df - di Really just tangents to the curve at a point. Velocity: AVERAGE Acceleration: INSTANTANEOUS

1 m/s -1 m/s 4 m/s +4 m/s 9 m/s +9 m/s 14 m/s +14 m/s 19 m/s +19 m/s

vi = 60 mi/h v2 = 0 mi/h a = -5 mi/h2 Call right + and left – vi = 5 m/s right = + 5 m/s vf = 4.8 m/s left = -4.8 m/s a = (vf – vi) / t = (-4.8 – 5) / .002sec = -4,900 m/s2 = 4,900 m/s2 left vi = 60 mi/h v2 = 0 mi/h a = -5 mi/h2 a = (vf – vi) / t  -5 = (0 – 60) / t  t = 12 s

x-t v-t END OF TODAY’S LECTURE

x-t ‘s x t UNIFORM Velocity Speed increases as slope increases x t Object at REST x t Moving forward or backward x t x-t ‘s Object Positively Accelerating x t x t Changing Direction x t Object Speeding up Object Negatively Accelerating

POSITIVE OR NEGATIVE ACCEL? SPEEDING UP OR SLOWING DOWN? x t x t Slope of the tangent gives vinst Getting more sloped  speeding up Getting more + sloped  + Accel Getting less sloped  slowing down Slopes are getting less +  - Accel x t x t Getting more sloped  speeding up Slopes are getting more –  - Accel Getting less sloped  slowing down Slopes are getting less –  + Accel

An easy way to remember it  I’m Positive!!! I’m Negative!!!

Find the acceleration in each case. v1 = 10 m/s, v2 = 20 m/s, Dt = 5sec v1 = 10 m/s, v2 = -20 m/s, Dt = 10sec v1 = -9 km/h, v2 = -27 km/h, Dt = 3 h v1 = -9 km/h, v2 = 6 km/h, Dt = 3 h

v-t ‘s UNIFORM Positive (+) Acceleration Acceleration increases as slope increases v t v t Changing Direction v-t ‘s UNIFORM Velocity (no acceleration) v t v t Object at REST UNIFORM Negative (-) Acceleration

v-t graphs v(m/s) Constant + accel (slowing down) At rest t (s) 2 4 6 8 10 12 t (s) 8 6 4 2 -2 -4 At rest Constant + accel (slowing down) Constant - Vel Constant negative accel (speeding up) Constant negative accel (slowing down) Constant + Vel (constant speed) Constant + accel (speeding up)

How to get the velocity (v) at a certain time off a v-t graph v(m/s) 2 4 6 8 10 12 t (s) 8 6 4 2 -2 -4 Example: What is the velocity at t = 8 seconds? Go over to t = 8. Find the pt on the graph. -2 m/s Find the v value for this time.

Finding the average acceleration on a v-t graph v(m/s) 2 4 6 8 10 12 8 6 4 2 -2 -4 Example: What is the average acceleration between 0 & 2, 2 & 4, and 4 & 10 seconds? a0-2 = (vf – vi) / Dt = rise / run = +4/2 = +2 m/s2 A2-4 = (vf – vi) / Dt = rise / run = 0 m/s2 A4-10 = (vf – vi) / Dt = rise / run = -7 / 6 = -1.17 m/s2

v-t graphs Slope of any segment is the AVERAGE acceleration v (m/s) The slope of the tangent to the curve at any point is the INSTANTANEOUS acceleration t (sec) t0 t1

Open to in your Unit 1 packet 3 30 20 10 -10 -20 -30 v = -30 m/s s = 30 m/s a = slope = (-30 m/s) / 16sec = -1.875 m/s2 a = slope = (+57 m/s) / 32sec = +1.78 m/s2 (approx)

Open to in your Unit 1 packet 3 30 20 10 -10 -20 -30 3) 4) 5) You can’t say. You know its speed at the start, but not where it is  Object is at rest whenever it crosses the t-axis  t = 0, 36, 80 sec Const – accel (object speeds up), const – vel, const + accel (slows down), const + accel (speeds up), const – accel (slows down) END OF LECTURE

x-t ‘s x t UNIFORM Velocity Speed increases as slope increases x t Object at REST x t Moving forward or backward x t x-t ‘s Object Positively Accelerating x t x t Changing Direction x t Object Speeding up Object Negatively Accelerating

v-t ‘s UNIFORM Positive (+) Acceleration Acceleration increases as slope increases v t v t Changing Direction v-t ‘s UNIFORM Velocity (no acceleration) v t v t Object at REST UNIFORM Negative (-) Acceleration

A Quick Review The slope between 2 points on an x-t graph gets you the _______________. The slope at a single point (the slope of the tangent to the curve) on an x-t graph gets you the ____________. The slope between 2 points on a v-t graph gets you the ____________. The slope at a single point (the slope of the tangent to the curve) on a v-t graph gets you the ____________. Average velocity Inst. velocity Avg. accel. Inst. accel.

NEW CONCEPT When you find the area “under the curve” on a v-t graph, this gets you the displacement during the given time interval. v t v t The “area under the curve” is really the area between the graph and the t-axis. This is NOT the area under the curve 

A = ½ (4)(-10) = -20m A = ½ (2)(-10) = -10m A = ½ (4)(15) = 30m Find the area under the curve from …. 0-4 seconds. 4-6 6-10 0-10 The displacement during the first 4 seconds is -20m A = ½ (4)(-10) = -20m A = ½ (2)(-10) = -10m The displacement during the next 2 seconds is -10m A = ½ (4)(15) = 30m The displacement during the next 4 seconds is 30m v t 15 -10 A = -20 + (-10) + 30 = 0m 4 6 10 The OVERALL displacement from 0 to 10 seconds is zero (its back to where it started)

How to find the displacement from one time to another from a v-t graph v(m/s) 2 4 6 8 10 12 t (s) 8 6 4 2 -2 -4 Example: What is the displacement from t = 2 to t = 10? Find the positive area bounded by the “curve” 12 m + 4 m = 16 m Find the negative area bounded by the “curve” (-2.25 m) + (-4.5 m) = - 6.75 m Add the positive and negative areas together  Dx = 16 m + (-6.75 m) = 9.25 m

How to find the distance traveled from one time to another from a v-t graph v(m/s) 2 4 6 8 10 12 t (s) 8 6 4 2 -2 -4 Example: What is the distance traveled from t = 2 to t = 10? Find the positive area bounded by the “curve” 12 m + 4 m = 16 m Find the negative area bounded by the “curve” (-2.25 m) + (-4.5 m) = - 6.75 m Add the MAGNITUDES of these two areas together  distance = 16 m + 6.75 m = 22.75 m

How to find the average velocity during a time interval on a v-t graph v(m/s) 2 4 6 8 10 12 t (s) 8 6 4 2 -2 -4 Example: What is the average velocity from t = 2 to t = 10? The DISPLACEMENT is simply the area “under” the curve. Dx = 16 m + (-6.75 m) = 9.25 m 12 m + 4 m = 16 m (-2.25 m) + (-4.5 m) = - 6.75 The AVG velocity = Dx / Dt = 9.25 m / 8 s = 1.22 m/s

How to find the final position of an object using a v-t graph (and being given the initial position) v(m/s) 2 4 6 8 10 12 t (s) 8 6 4 2 -2 -4 Example: What is the final position after t = 10 seconds if xi = 40 m? The DISPLACEMENT during the 1st 10 sec is simply the area “under” the curve. Dx = 20 m + (-6.75 m) = 13.25 m 4 m 12 m + 4 m = 20 m (-2.25 m) + (-4.5 m) = - 6.75 Dx = xf – xi  xf = Dx + xi = 13.25 m + 40 m = 53.25 m

v-t graphs Slope of any segment is the AVERAGE acceleration v (m/s) The slope of the tangent to the curve at any point is the INSTANTANEOUS acceleration The area under the curve between any two times is the CHANGE in position (the displacement) during that time period. t (sec) t0 t1

Open to in your Unit 1 packet 3 30 20 10 -10 -20 -30 6) 7) Distance travelled = |area| = | ½ (16)(-30) | = 240m s = d/t = 240 m / 16 sec = 15 m/s Displacement = |area| = ½ (16)(-30) + 8 (-30) = -480m v = Dx/t = -480 m / 24 sec = -20 m/s

Open to in your Unit 1 packet 3 30 20 10 -10 -20 -30 8) Find all the areas “under the curve” from 0 to 44 sec Area = ½ (16)(-30) + 12(-30) + ½ (8)(-30) + ½ (8)(30) = - 600 m Area = Dx = - 600 m  Dx = xf – xi  -600m = xf – (-16m)  xf = - 616m

Open to in your Unit 1 packet 4 +3.3 m/s2 +10 m/s 0 m/s +75 m

10 -10 5 -2 m/s2 0 m 50 m 30 m

14 & 34 sec +.35 m/s2 + 8 m/s + 2 m/s2 approx 0.8 m/s2 9) 10) 11) 12) 13) 14 & 34 sec +.35 m/s2 + 8 m/s + 2 m/s2 approx 0.8 m/s2