![KINEMATICS DERIVATIONS (CONT.)
2
1
vf = v0 + at t = (vf – v0)/a
x = v0t + at2
x =v0 [(vf – v0)/a] + a[(vf – v0)/a]2
vf
2
– v0
2
= 2ax
2
1
Note that the top equation is solved for t and that
expression for t is substituted twice (in red) into the
x equation. You should work out the algebra to prove
the final result on the last line.](https://image.slidesharecdn.com/11thphysicsunit2introduction-250213040817-761c76df/85/11th-physics-unit-2-introduction-to-current-voltage-18-320.jpg)
11th physics unit 2 introduction to current voltage
- 1. PHYSICS INTRO & KINEMATICS - BY VIDHYA BASKARAN
- 2. PHYSICS INTRO & KINEMATICS •Quantities •Units •Vectors •Displacement •Velocity •Acceleration •Kinematics •Graphing Motion in 1-D
- 3. SOME PHYSICS QUANTITIES Vector - quantity with both magnitude (size) and direction Scalar - quantity with magnitude only Vectors: • Displacement • Velocity • Acceleration • Momentum • Force Scalars: • Distance • Speed • Time • Mass • Energy
- 4. MASS VS. WEIGHT On the moon, your mass would be the same, but the magnitude of your weight would be less. Mass • Scalar (no direction) • Measures the amount of matter in an object Weight • Vector (points toward center of Earth) • Force of gravity on an object
- 5. VECTORS • THE LENGTH OF THE ARROW REPRESENTS THE MAGNITUDE (HOW FAR, HOW FAST, HOW STRONG, ETC, DEPENDING ON THE TYPE OF VECTOR). • THE ARROW POINTS IN THE DIRECTIONS OF THE FORCE, MOTION, DISPLACEMENT, ETC. IT IS OFTEN SPECIFIED BY AN ANGLE. Vectors are represented with arrows 42° 5 m/s
- 6. UNITS QUANTITY . . . UNIT (SYMBOL) • DISPLACEMENT & DISTANCE . . . METER (M) • TIME . . . SECOND (S) • VELOCITY & SPEED . . . (M/S) • ACCELERATION . . . (M/S2 ) • MASS . . . KILOGRAM (KG) • MOMENTUM . . . (KG·M/S) • FORCE . . .NEWTON (N) • ENERGY . . . JOULE (J) Units are not the same as quantities!
- 7. SI PREFIXES pico p 10-12 nano n 10-9 micro µ 10-6 milli m 10-3 centi c 10-2 kilo k 103 mega M 106 giga G 109 tera T 1012 Little Guys Big Guys
- 8. KINEMATICS DEFINITIONS • KINEMATICS – BRANCH OF PHYSICS; STUDY OF MOTION • POSITION (X) – WHERE YOU ARE LOCATED • DISTANCE (D ) – HOW FAR YOU HAVE TRAVELED, REGARDLESS OF DIRECTION • DISPLACEMENT (X) – WHERE YOU ARE IN RELATION TO WHERE YOU STARTED
- 9. DISTANCE VS. DISPLACEMENT • YOU DRIVE THE PATH, AND YOUR ODOMETER GOES UP BY 8 MILES (YOUR DISTANCE). • YOUR DISPLACEMENT IS THE SHORTER DIRECTED DISTANCE FROM START TO STOP (GREEN ARROW). • WHAT IF YOU DROVE IN A CIRCLE? start stop
- 10. SPEED, VELOCITY, & ACCELERATION • SPEED (V) – HOW FAST YOU GO • VELOCITY (V) – HOW FAST AND WHICH WAY; THE RATE AT WHICH POSITION CHANGES • AVERAGE SPEED ( V ) – DISTANCE/TIME • ACCELERATION (A) – HOW FAST YOU SPEED UP, SLOW DOWN, OR CHANGE DIRECTION; THE RATE AT WHICH VELOCITY CHANGES
- 11. SPEED VS. VELOCITY • SPEED IS A SCALAR (HOW FAST SOMETHING IS MOVING REGARDLESS OF ITS DIRECTION). EX: V = 20 MPH • SPEED IS THE MAGNITUDE OF VELOCITY. • VELOCITY IS A COMBINATION OF SPEED AND DIRECTION. EX: V = 20 MPH AT 15 SOUTH OF WEST • THE SYMBOL FOR SPEED IS V. • THE SYMBOL FOR VELOCITY IS TYPE WRITTEN IN BOLD: V OR HAND WRITTEN WITH AN ARROW: V
- 12. SPEED VS. VELOCITY • DURING YOUR 8 MI. TRIP, WHICH TOOK 15 MIN., YOUR SPEEDOMETER DISPLAYS YOUR INSTANTANEOUS SPEED, WHICH VARIES THROUGHOUT THE TRIP. • YOUR AVERAGE SPEED IS 32 MI/HR. • YOUR AVERAGE VELOCITY IS 32 MI/HR IN A SE DIRECTION. • AT ANY POINT IN TIME, YOUR VELOCITY VECTOR POINTS TANGENT TO YOUR PATH. • THE FASTER YOU GO, THE LONGER YOUR VELOCITY VECTOR.
- 13. ACCELERATION ACCELERATION – HOW FAST YOU SPEED UP, SLOW DOWN, OR CHANGE DIRECTION; IT’S THE RATE AT WHICH VELOCITY CHANGES. TWO EXAMPLES: t (s) v (mph) 0 55 1 57 2 59 3 61 t (s) v (m/s) 0 34 1 31 2 28 3 25 a = +2 mph/s a = -3 m/s s = -3 m/s2
- 14. VELOCITY & ACCELERATION SIGN CHART V E L O C I T Y A C C E L E R A T I O N + - + Moving forward; Speeding up Moving backward; Slowing down - Moving forward; Slowing down Moving backward; Speeding up
- 15. ACCELERATION DUE TO GRAVITY A = -G = -9.8 M/S2 9.8 m/s2 Near the surface of the Earth, all objects accelerate at the same rate (ignoring air resistance). Interpretation: Velocity decreases by 9.8 m/s each second, meaning velocity is becoming less positive or more negative. Less positive means slowing down while going up. More negative means speeding up while going down. This acceleration vector is the same on the way up, at the top, and on the way down!
- 16. KINEMATICS FORMULA SUMMARY (derivations to follow) • vf = v0 + at • vavg = (v0 + vf )/2 x = v0t + ½ at2 • vf 2 – v0 2 = 2 ax 2 1 For 1-D motion with constant acceleration:
- 17. KINEMATICS DERIVATIONS a = v/t (by definition) a = (vf – v0)/t vf = v0 + at vavg = (v0 + vf )/2 will be proven when we do graphing. x = vt = ½ (v0 + vf)t = ½ (v0 + v0 + at)t x = v0 t + at2 2 1 (cont.)
- 18. KINEMATICS DERIVATIONS (CONT.) 2 1 vf = v0 + at t = (vf – v0)/a x = v0t + at2 x =v0 [(vf – v0)/a] + a[(vf – v0)/a]2 vf 2 – v0 2 = 2ax 2 1 Note that the top equation is solved for t and that expression for t is substituted twice (in red) into the x equation. You should work out the algebra to prove the final result on the last line.
- 19. SAMPLE PROBLEMS 1. YOU’RE RIDING A UNICORN AT 25 M/S AND COME TO A UNIFORM STOP AT A RED LIGHT 20 M AWAY. WHAT’S YOUR ACCELERATION? 2. A BRICK IS DROPPED FROM 100 M UP. FIND ITS IMPACT VELOCITY AND AIR TIME. 3. AN ARROW IS SHOT STRAIGHT UP FROM A PIT 12 M BELOW GROUND AT 38 M/S. a. FIND ITS MAX HEIGHT ABOVE GROUND. b. AT WHAT TIMES IS IT AT GROUND LEVEL?
- 20. MULTI-STEP PROBLEMS 1. HOW FAST SHOULD YOU THROW A KUMQUAT STRAIGHT DOWN FROM 40 M UP SO THAT ITS IMPACT SPEED WOULD BE THE SAME AS A MANGO’S DROPPED FROM 60 M? 2. A DUNE BUGGY ACCELERATES UNIFORMLY AT 1.5 M/S2 FROM REST TO 22 M/S. THEN THE BRAKES ARE APPLIED AND IT STOPS 2.5 S LATER. FIND THE TOTAL DISTANCE TRAVELED. 19.8 m/s 188.83 m Answer: Answer:
- 21. GRAPHING ! x t A B C A … Starts at home (origin) and goes forward slowly B … Not moving (position remains constant as time progresses) C … Turns around and goes in the other direction quickly, passing up home 1 – D Motion
- 22. GRAPHING W/ ACCELERATION x A … Start from rest south of home; increase speed gradually B … Pass home; gradually slow to a stop (still moving north) C … Turn around; gradually speed back up again heading south D … Continue heading south; gradually slow to a stop near the starting point t A B C D
- 23. TANGENT LINES t SLOPE VELOCITY Positive Positive Negative Negative Zero Zero SLOPE SPEED Steep Fast Gentle Slow Flat Zero x On a position vs. time graph:
- 24. INCREASING & DECREASING t x Increasing Decreasing On a position vs. time graph: Increasing means moving forward (positive direction). Decreasing means moving backwards (negative direction).
- 25. CONCAVITY t x On a position vs. time graph: Concave up means positive acceleration. Concave down means negative acceleration.
- 26. SPECIAL POINTS t x P Q R Inflection Pt. P, R Change of concavity Peak or Valley Q Turning point Time Axis Intercept P, S Times when you are at “home” S
- 27. CURVE SUMMARY t x Concave Up Concave Down Increasing v > 0 a > 0 (A) v > 0 a < 0 (B) Decreasing v < 0 a > 0 (D) v < 0 a < 0 (C) A B C D
- 29. GRAPHING ANIMATION LINK This website will allow you to set the initial velocity and acceleration of a car. As the car moves, all three graphs are generated. Car Animation
- 30. GRAPHING TIPS • Line up the graphs vertically. • Draw vertical dashed lines at special points except intercepts. • Map the slopes of the position graph onto the velocity graph. • A red peak or valley means a blue time intercept. t x v t
- 31. GRAPHING TIPS The same rules apply in making an acceleration graph from a velocity graph. Just graph the slopes! Note: a positive constant slope in blue means a positive constant green segment. The steeper the blue slope, the farther the green segment is from the time axis. a t v t
- 32. REAL LIFE Note how the v graph is pointy and the a graph skips. In real life, the blue points would be smooth curves and the green segments would be connected. In our class, however, we’ll mainly deal with constant acceleration. a t v t
- 33. AREA UNDER A VELOCITY GRAPH v t “forward area” “backward area” Area above the time axis = forward (positive) displacement. Area below the time axis = backward (negative) displacement. Net area (above - below) = net displacement. Total area (above + below) = total distance traveled.
- 34. AREA The areas above and below are about equal, so even though a significant distance may have been covered, the displacement is about zero, meaning the stopping point was near the starting point. The position graph shows this too. v t “forward area” “backward area” t x
- 35. AREA UNITS • IMAGINE APPROXIMATING THE AREA UNDER THE CURVE WITH VERY THIN RECTANGLES. • EACH HAS AREA OF HEIGHT WIDTH. • THE HEIGHT IS IN M/S; WIDTH IS IN SECONDS. • THEREFORE, AREA IS IN METERS! v (m/s) t (s) 12 m/s 0.5 s 12 • The rectangles under the time axis have negative
- 36. GRAPHS OF A BALL THROWN STRAIGHT UP x v a The ball is thrown from the ground, and it lands on a ledge. The position graph is parabolic. The ball peaks at the parabola’s vertex. The v graph has a slope of -9.8 m/s2 . Map out the slopes! There is more “positive area” than negative on the v graph. t t t
- 37. GRAPH PRACTICE Try making all three graphs for the following scenario: 1. Schmedrick starts out north of home. At time zero he’s driving a cement mixer south very fast at a constant speed. 2. He accidentally runs over an innocent moose crossing the road, so he slows to a stop to check on the poor moose. 3. He pauses for a while until he determines the moose is squashed flat and deader than a doornail. 4. Fleeing the scene of the crime, Schmedrick takes off again in the same direction, speeding up quickly. 5. When his conscience gets the better of him, he slows, turns around, and returns to the crash site.
- 38. KINEMATICS PRACTICE A catcher catches a 90 mph fast ball. His glove compresses 4.5 cm. How long does it take to come to a complete stop? Be mindful of your units! 2.24 ms Answer
- 39. UNIFORM ACCELERATION WHEN OBJECT STARTS FROM REST AND UNDERGOES CONSTANT ACCELERATION: • POSITION IS PROPORTIONAL TO THE SQUARE OF TIME. • POSITION CHANGES RESULT IN THE SEQUENCE OF ODD NUMBERS. • FALLING BODIES EXHIBIT THIS TYPE OF MOTION (SINCE G IS CONSTANT). t : 0 1 2 3 4 x = 1 x = 3 x = 5 ( arbitrary units ) x : 0 1 4 9 16 x = 7
- 40. SPREADSHEET PROBLEM • WE’RE ANALYZING POSITION AS A FUNCTION OF TIME, INITIAL VELOCITY, AND CONSTANT ACCELERATION. • X, X, AND THE RATIO DEPEND ON T, V0, AND A. • X IS HOW MUCH POSITION CHANGES EACH SECOND. • THE RATIO (1, 3, 5, 7) IS THE RATIO OF THE X’S. t (s) x (m) delta x (m) ratio v 0 (m/s) a (m/s2 ) 0 0 0 17.3 1 8.66 8.66 1 2 34.64 25.98 3 3 77.94 43.30 5 4 138.56 60.62 7 • Make a spreadsheet like this and determine what must be true about v0 and/or a in order to get this ratio of odd numbers. • Explain your answer mathematically.
- 41. RELATIONSHIPS Let’s use the kinematics equations to answer these: 1. A mango is dropped from a height h. a. If dropped from a height of 2h, would the impact speed double? b. Would the air time double when dropped from a height of 2h ? 2. A mango is thrown down at a speed v. a. If thrown down at 2v from the same height, would the impact speed double? b. Would the air time double in this case?
- 42. RELATIONSHIPS (CONT.) 3. A RUBBER CHICKEN IS LAUNCHED STRAIGHT UP AT SPEED V FROM GROUND LEVEL. FIND EACH OF THE FOLLOWING IF THE LAUNCH SPEED IS TRIPLED (IN TERMS OF ANY CONSTANTS AND V). a. MAX HEIGHT b. HANG TIME c. IMPACT SPEED 3v 9v2 /2g 6v/g Answers