unit 1 motion graphs. - mechanics kinematics dynamics - electricity - magnetism - optics - waves 5...

Unit 1 Motion Graphs

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

The purpose of this chapter is to learn the 1 st 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 (20C) 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 objects path is straight, the distance is the length of the straight line between start and finish. ** If an objects 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

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

Displacement the change in an objects 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 doesnt 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).

Using the number line above, find the distance travelled and the displacement in moving from - A to B - C to A - A to C and then back to A - C to B, passing through A BC -3 -2 -1 0 1 A meters 1m, 1m [right]* *notice a direction is given for displacement 4m, 4m [left]* 8m, 0m no direction needed here 4m, 3m [left]* x = 1 (1m) = 0m x = (-2) (1m) = -3m OR 3m [left] + or can be used instead of R and L

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

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? BC -3 -2 -1 0 1 A miles 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

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 BC -3 -2 -1 0 1 A km a)Larrys average speed during the first portion of the trip. b)The average speed during the second portion of the trip. c)Larrys 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

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/ t (or v=d/t) *This line means its a vector

AD 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 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 Find Sams avg. speed and avg. velocity for the entire trip. Avg Velocity = 0 since x = 0 for the entire trip. He ended in the exact location he started!!

AD 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 31.831 m 100 104.94 Find Sams average speed and average velocity for the 1 st half of the race. s = d/t t = d/s = 200m/(14.286+12.5s) = 7.47 m/s v = x/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? distancespeedtime What vectors have we learned about thus far? Displacementvelocity

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

Scalars vs. Vectors (continued) has magnitude & direction (example: 15 mi/h North) has a magnitude only (example: 30 km/h) If an object STARTS & STOPS at the same point, the velocity is ZERO! (since the displacement is zero) Velocity: Speed: 1 2 24 km 7 km Total time for the trip from 1 to 2: 1 hr Speed = 31 km/h Velocity = 25 km/h at 16 o NE 25 km 16 o Dont worry about this notation for this test just give the general direction

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. SLOPE s AB = rise/run = (30-0m) / (10-0s) = 3 m/s s CD = rise/run = (100-50m) / (20-15s) = 10 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

d-t graphs (distance-time) Constant speed Speeding UP notice how more distance is covered each second Constant Speed (but faster than AB) Slowing Downless 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

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 Practice Graph

x-t graphs (position time graphs. Like d-t graphs) t (sec) x (m) t 1 t 2 t 3 x2x1x3x2x1x3 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 ( displacement)

How to get the displacement/position (d) at a certain time (t) off an d-t graph. Sometimes Ill refer to this as x- t d(m) 10 20 30 40 50 t (s) 30 20 10 0 Example: What is the position at t = 30 seconds? Go over to t = 30. Find the pt on the curve. Find the x value for this time. 24m

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 0 Example: What is the displacement from t = 10 to t = 40? Find x i Find x f Use x = x f - x i = + 7 m 10 m 17 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 0 Example: What is the distance traveled from t = 10 to t = 40? 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) 17 m 10 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 0 Example: What is the average speed from t = 10 to t = 40 seconds? dist 10-40 = 27 m (previous slide) Avg. Speed = dist/ t = 27m / 30 sec = 0.9 m/s 17 m 10 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 0 Example: What is the average velocity from t = 10 to t = 40 seconds? Avg. Velocity = slope = x/ t = + 7 / 30 sec = + 0.23 m/s x 10-40 = + 7 m (previous slide) 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 0 Avg. vel. = slope = rise/run = -7 m / 20 = -.35 m/s Example: What is the average velocity from t = 20 to t = 40 seconds? Since the objects displacement is in the NEGATIVE direction, so is its average velocity.

-10 m avg velocity = slope = -15m / 6sec = -2.5 m/s s = |v| = 2.5 m/s At rest at t = 0 and t = 12 sec Graph Example 2) 3) 4)

x = x 2 x 1 = (-10m) (10m) = -20m (approximately) 5) 6) Speeding up, const negative vel, slowing down, speeding up, const positive velocity(slow), speeding up, constant positive velocity (fast)

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

Just what is meant by instantaneous velocity? tt tt tt tt tt 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. 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

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 0 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) Example: What is the instantaneous velocity at t = 20 seconds? YOU MUST SPECIFY WHICH POINTS YOU USED WHEN FINDING THE SLOPE!!!! (24, 30) (15, 15)

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 0 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 (0,5) (10,10) 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 0 Draw the tangent to the curve at the point in question. Then, find the slope of the tangent. Slope = 0 (object at rest) Example: What is the instantaneous velocity at t = 25 seconds?

x-t graphs t (sec) x (m) t 1 t 2 t 3 x2x1x3x2x1x3 1 2 3 0 Slope of line segment

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

Definition Average Acceleration ( a ) the change in an objects velocity in a given time interval..IN OTHER WORDS, the rate of change of an objects 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 = v f -v iOR a = v/t t f -t i

Find the acceleration of each object 1) An object is moving at 40 m/s when it slows down to 20 m/s over a 10 second interval. 2) An object is moving at -40 m/s, and 5 seconds earlier it was moving at -10 m/s. 3) 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. a = v/ t = (20 40m/s) / 10sec = -2 m/s 2 a = v/ t = [-40 (-10m/s)] / 5sec = -6 m/s 2 a = v/ t = [20 (-10 in/min)] / 2min = +15 in/min 2 Slows down Negative accel Speeds up Negative accel Speeds up + Accel a = v/ t = [-20 (-30 mi/h)] / 10hr = + 1mi/h 2 Slows down + Accel

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 doesnt 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? 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, v is +

What do the units of acceleration mean? m/s 2 3 m/s 2 means that your velocity increased by 3 m/s every second. -0.1 km/min 2 means that your velocity decreased by 0.1 km/min every minute that you were moving. m/s/s m/s per second t (sec)v (m/s) 00 13 26 39 412 t (min)v (km/min) 00.3 10.2 20.1 30 4-0.1

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

-1 m/s1 m/s +4 m/s4 m/s +9 m/s9 m/s +14 m/s14 m/s +19 m/s19 m/s

Call right + and left v i = 5 m/s right = + 5 m/sv f = 4.8 m/s left = -4.8 m/s a = (v f v i ) / t = (-4.8 5) /.002sec = -4,900 m/s 2 = 4,900 m/s 2 left v i = 60 mi/hv 2 = 0 mi/ha = -5 mi/h 2 a = (v f v i ) / t -5 = (0 60) / t t = 12 s

x-t v-t END OF TODAYS LECTURE

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

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

An easy way to remember it Im Negative!!! Im Positive!!!

Find the acceleration in each case. v 1 = 10 m/s, v 2 = 20 m/s, t = 5sec v 1 = 10 m/s, v 2 = -20 m/s, t = 10sec v 1 = -9 km/h, v 2 = -27 km/h, t = 3 h v 1 = -9 km/h, v 2 = 6 km/h, t = 3 h

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

v(m/s) 2 4 6 8 10 12 t (s) 8 6 4 2 0 -2 -4 v-t graphs Constant + accel (speeding up) Constant + Vel (constant speed) Constant negative accel (slowing down) At rest Constant negative accel (speeding up) Constant + accel (slowing down) Constant - Vel

v(m/s) 2 4 6 8 10 12 t (s) 8 6 4 2 0 -2 -4 How to get the velocity (v) at a certain time off a v-t graph Example: What is the velocity at t = 8 seconds? Go over to t = 8. Find the pt on the graph. Find the v value for this time. -2 m/s

v(m/s) 2 4 6 8 10 12 8 6 4 2 0 -2 -4 Finding the average acceleration on a v-t graph A 2-4 = (v f v i ) / t = rise / run = 0 m/s 2 A 4-10 = (v f v i ) / t = rise / run = -7 / 6 = -1.17 m/s 2 Example: What is the average acceleration between 0 & 2, 2 & 4, and 4 & 10 seconds? a 0-2 = (v f v i ) / t = rise / run = +4/2 = +2 m/s 2

v-t graphs t (sec) v (m/s) Slope of any segment is the AVERAGE acceleration The slope of the tangent to the curve at any point is the INSTANTANEOUS acceleration t 0 t 1

30 20 10 0 -10 -20 -30 a = slope = (+57 m/s) / 32sec = +1.78 m/s 2 (approx) Open to in your Unit 1 packet 3 v = -30 m/ss = 30 m/s a = slope = (-30 m/s) / 16sec = -1.875 m/s 2

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

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

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

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 Inst. accel. Avg. 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. This is NOT the area under the curve The area under the curve is really the area between the graph and the t-axis. v t v t

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

v(m/s) 24 6 8 10 12 t (s) 8 6 4 2 0 -2 -4 How to find the displacement from one time to another from a v-t graph Example: What is the displacement from t = 2 to t = 10? Add the positive and negative areas together x = 16 m + (-6.75 m) = 9.25 m 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

v(m/s) 24 6 8 10 12 t (s) 8 6 4 2 0 -2 -4 How to find the distance traveled from one time to another from a v-t graph Example: What is the distance traveled from t = 2 to t = 10? Add the MAGNITUDES of these two areas together distance = 16 m + 6.75 m = 22.75 m 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

v(m/s) 24 6 8 10 12 t (s) 8 6 4 2 0 -2 -4 How to find the average velocity during a time interval on a v-t graph Example: What is the average velocity from t = 2 to t = 10? The AVG velocity = x / t = 9.25 m / 8 s = 1.22 m/s The DISPLACEMENT is simply the area under the curve. x = 16 m + (-6.75 m) = 9.25 m 12 m + 4 m = 16 m (-2.25 m) + (-4.5 m) = - 6.75

v(m/s) 24 6 8 10 12 t (s) 8 6 4 2 0 -2 -4 How to find the final position of an object using a v-t graph (and being given the initial position) Example: What is the final position after t = 10 seconds if x i = 40 m? x = x f x i x f = x + x i = 13.25 m + 40 m = 53.25 m The DISPLACEMENT during the 1 st 10 sec is simply the area under the curve. x = 20 m + (-6.75 m) = 13.25 m 4 m 12 m + 4 m = 20 m (-2.25 m) + (-4.5 m) = - 6.75

v-t graphs t (sec) v (m/s) Slope of any segment is the AVERAGE acceleration The slope of the tangent to the curve at any point is the INSTANTANEOUS acceleration t 0 t 1 The area under the curve between any two times is the CHANGE in position (the displacement) during that time period.

Open to in your Unit 1 packet 3 Area = (16)(-30) + 12(-30) + (8)(-30) + (8)(30) = - 600 m 8) Find all the areas under the curve from 0 to 44 sec Area = x = - 600 m x = x f x i -600m = x f (-16m) x f = - 616m 30 20 10 0 -10 -20 -30

Open to in your Unit 1 packet 4 +3.3 m/s 2 +10 m/s 0 m/s +75 m

-2 m/s 2 10 -10 0 m 50 m30 m 5

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

unit 1 motion graphs. - mechanics kinematics dynamics - electricity - magnetism - optics - waves 5...

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