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# Mechanics

## on May 11, 2010

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A set of slides created to teach Mechanics to learners at Bishops Diocesan College in Cape Town.

A set of slides created to teach Mechanics to learners at Bishops Diocesan College in Cape Town.

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## MechanicsPresentation Transcript

• Position?
• Position of an object is always defined ………. to other objects.
• Compass points.
• Absolute bearings. (From …………..) always taken to the ……….
? … .. … .. S N NE …… o …… . …… . … . … . o …… o …… o …… .. o … .. …… . o 360 o /0
• Position?
• Position of an object is always defined relative to other objects.
• Compass points.
• Absolute bearings. (From North) always taken to the right.
? E W S N NE 45 o NW SW SE 90 o 135 o 180 o 225 o 270 o 315 o 360 o /0
• Displacement & Distance A B Distance = ………………. Home School Straight Line distance = …………………. Displacement = ……………………………….
• Displacement & Distance A B Distance = Path traveled Home School Straight Line distance = displacement Displacement = distance Displacement is a VECTOR – it has both MAGNITUDE and DIRECTION
• Negative Displacement Displacement (decreasing) = distance increasing (positive) Displacement – S = 1. Moving back to start. 2. Moving away in reverse direction.
• Negative Displacement Displacement ( S ) (decreasing) = distance increasing (positive) Displacement – negative increasing = distance increasing ( positive )  S = s 2 - s 1 = 2 - 6 = -4 m Change in displacement 4m CLOSER to starting pt. NEGATIVE DIRECTION. 1. Moving back to start. (Choose Right as positive.) 2. Moving away in reverse direction. s 1 = 6m S 2 = 2m +  S  S = s 2 - s 1 = - 2 - 6 = - 8 m Change in displacement 8m in negative direction s 1 = 6m S 2 = -2m  S
• Units of Distance & Displacement
• K m H m D m m d m c m mm
1 m Dm Hm Km 1000 (10 3 ) 100 (10 2 ) 10 1 / 10 (10 -1 ) 1 / 100 (10 -2 ) 1 / 1000 (10 -3 ) dm cm mm 1m = ……Dm = ……..Hm = ……..Km 1m = ….dm = …….cm = ………..mm
• Units of Distance & Displacement
• K m H m D m m d m c m mm
1 m 0. 1 Dm 0. 0 1 Hm 0. 0 0 1 Km 1000 (10 3 ) 100 (10 2 ) 10 1 / 10 (10 -1 ) 1 / 100 (10 -2 ) 1 / 1000 (10 -3 ) 1 0 dm 1 0 0 cm 1 0 0 0 mm 1m = 0.1 Dm = 0.01Hm = 0.001Km 1m = 10dm = 100cm = 1000mm
• Units of Distance & Displacement
• Convert the following to m.
• 5 cm
• 20 mm
• 40 km
• 2 dm
• 10 Hm
• 0.1 cm
• 0.11 dm
• 0.506 Dm
• Length of your foot (m)
• Lab dimensions (mm)
• Units of Distance & Displacement
• Convert the following to m.
• 5 cm = 0.05 m
• 20 mm = 0.02 m
• 40 km = 40 000 m
• 2 dm = 0.2 m
• 10 Hm = 1000 m
• 0.1 cm = 0.001 m
• 0.11 dm = 0.011 m
• 0.506 Dm = 5.06 m
• Foot 30.5 cm = 0.305 m
• Height (km)1.75 m = 0.00175 Km
• Lab length (mm) = 15 500 mm
• Speed & Velocity * A * B Path traveled Position (distance & displacement) is changing by the _________________ every second. |        Uniform velocity s … . … . … . … . … . … . … . … . Average Speed = … . … . Average Velocity = ………… . ………… .. V = = V = =
• Speed & Velocity |        * A * B Path traveled 70Km Position (distance & displacement) is changing by the same amount every second. Uniform velocity Displacement = 50Km T = 24 hrs Average Speed = Distance traveled time Average Velocity = Displacement time V = = = = 0.81 m.s -1 ∆ s s 2 -s 1 70 000 ∆ t t 2 -t 1 86400 V = = = = 0.58 m.s -1 towards B ∆ s s 2 -s 1 50 000 ∆ t t 2 -t 1 86400
• Negative Velocity Displacement …...: Velocity ……. (positive - …………..) Displacement – _________________ increasing = distance increasing ( ____________ ) V = = …. ∆ s ∆ t - + V = ∆ s ∆ t - + Displacement < 0 .: V < 0 (_______) 1. Moving back to start 2. In reverse direction from starting point S 1 S 2
• Negative Velocity Displacement < 0 .: Velocity < 0 (positive - decreasing) Displacement – negative increasing = distance increasing ( positive ) V = = - ∆ s ∆ t - + V = ∆ s ∆ t - + Displacement < 0 .: V < 0 (NEGATIVE) 1. Moving back to start (RHS = positive – decision) 2. In reverse direction from starting point
• Units of Velocity
• Convert between different units of speed and velocity, e.g. 60 km.h -1  ?? m.s -1 ,
• Units of Velocity
• Convert between different units of speed and velocity, e.g. m.s -1 , km.h -1 .
60 m.s -1  0.06Km.s -1  0.06x60x60 = 216 Km/h 10 m.s -1 = 36 Km/h
• Example: An athlete runs one lap of a 400m track A-A in 60s at constant (uniform) speed. Find:- 1. distance A-B 2. displacement A-B 3. average speed A-B 4. average velocity A-B 5. Average velocity A-A A B
• Example:
• An athlete runs one lap A-A in 60s at constant (uniform) speed.
• distance A-B = 200m
• displacement A-B
• Circum circle = 2  r .: 200 =  d .: d = 200/  = 63.66 m
• S(A-B) =  (63.66) 2 + (100) 2 =  14 052.85 m = 118.5 m towards B
• average speed A-B
• v = s/t = 200/30 = 6.67 m.s -1
• average velocity A-B
• v = s/t = 118.5/30 = 3.95 m.s -1 towards B
• Average velocity A-A
• v = s/t = 0/60 = 0
A B
• Changing Velocity - Acceleration .. . . . . . . . . . . . . 0m/s 120Km/h 0 km/h 120 km/h Average = ??
• Changing Velocity - Acceleration
• The ………………….. VELOCITY at any point is …………………………… the INSTANTANEOUS VELOCITY.
• The AVERAGE VELOCITY over an ……………………… is ……………….. to the INSTANTANEOUS VELOCITY at the ……………………… of the interval.
.. . . . . . . . . . . . . 0m/s 2m/s 4m/s 6m/s 8m/s 0 1s 2s 3s 4s TIME t (s) 0 1 2 3 4 5 6 TOT Total Time INSTANTANEOUS VELOCITY Inst v (m/s) 0 2 4 6 8 10 12 AVERAGE VELOCITY in EACH SECOND av v (m/s) 0 Average Velocity DISPLACEMENT s (m) 0 Total displacement EQUIVALENT UNIFORM VELOCITY Unfm v 0 Average Velocity
• Changing Velocity - Acceleration
• The AVERAGE VELOCITY at any point is HALF the INSTANTANEOUS VELOCITY .
• The AVERAGE VELOCITY over an interval is EQUAL to the INSTANTANEOUS VELOCITY at the midpoint of the interval.
.. . . . . . . . . . . . . 0m/s 2m/s 4m/s 6m/s 8m/s 0 1s 2s 3s 4s Rate of change in velocity = 2m.s -1 per second = acceleration! TIME t (s) 0 1 2 3 4 5 6 TOT Total Time INSTANTANEOUS VELOCITY Inst v (m/s) 0 2 4 6 8 10 12   12 INSTANTANEOUS VELOCITY AVERAGE VELOCITY /s av v (m/s) 0 1 2 3 4 5 6 6 Average Velocity DISPLACEMENT s (m) 0 1 4 9 16 25 36 36 Total displacement EQUIVALENT UNIFORM VELOCITY Unfm v 0 6 6 6 6 6 6 6 Average Velocity
• Changing Velocity - Acceleration ……………… Velocity ………… Velocity 14 12 10 8 6 4 2 0 TIME t (s) 0 1 2 3 4 5 6 6 Total Time INSTANTANEOUS VELOCITY Inst v (m/s) 0 2 4 6 8 10 12   AVERAGE VELOCITY /s av v (m/s) 0 1 2 3 4 5 6 6 Average Velocity DISPLACEMENT s (m) 0 1 4 9 16 25 36 36 Total displacement EQUIVALENT UNIFORM VELOCITY Unfm v 0 6 6 6 6 6 6 6 Average Velocity
• Changing Velocity - Acceleration
• The AVERAGE VELOCITY at any point is HALF the INSTANTANEOUS VELOCITY .
• The AVERAGE VELOCITY over an interval is EQUAL to the INSTANTANEOUS VELOCITY at the midpoint of the interval .
Instantaneous Velocity Average Velocity 14 12 10 8 6 4 2 0
• Acceleration
• Velocity ……………… – you travel ……………… every second.
• Acceleration is the ……… of ……………. of velocity .
• In SI units , acceleration is measured in ………………….(………).
Positive acceleration A > 0 Velocity increasing |        V 1 = 1m.s -1 V 2 = 4m.s -1 a = 2s
• Acceleration
• Velocity increases – you travel FURTHER every second.
• Acceleration is the rate of change of velocity .
• In SI units , acceleration is measured in metres/second² (m·s - ²).
Positive acceleration a > 0 Velocity increasing |        V 1 = 1m.s -1 V 2 = 4m.s -1 a = = = = 1.5 m.s -2 Δ v v 2 – v 1 4 – 1 Δ t t 2 – t 1 2 - 0 2s
• Deceleration
• Velocity ………………. – you travel ……………….. distance every second.
• Acceleration is the rate of change of velocity .
…………… . acceleration a 0 Velocity …………… . |       a = 4m.s -1 0m.s -1 2s
• Deceleration
• Velocity decrease – you travel shorter distance every second.
• Acceleration is the rate of change of velocity .
• In SI units , acceleration is measured in metres/second² (m·s-²).
Negative acceleration a < 0 Velocity decreasing . |       a = = = = - 2 m.s -2 Δ v v 2 – v 1 0 – 4 Δ t t 2 – t 1 2 - 0 4m.s -1 0m.s -1 2s
• Uniform acceleration For . Acceleration = rate of change of velocity a = V t An object that falls from rest. Time (s) Displacement (m) Average Velocity Instantaneous Velocity (m.s -1 ) Acceleration (m.s -2 ) 0 0 5 20 45 4 80 5 125 6 180 1 2 3
• Uniform acceleration For . Acceleration = rate of change of velocity a = V t Time (s) Displacement (m) Average Velocity Instantaneous Velocity (m.s-1) Acceleration (m.s-2) 0 0 0 5 5 20 10 45 15 4 80 20 5 125 25 6 180 30 1 2 3
• Uniform acceleration For . Acceleration = rate of change of velocity a = V t Time (s) Displacement (m) Average Velocity Instantaneous Velocity (m.s-1) Acceleration (m.s-2) 0 0 0 0 5 5 10 20 10 20 45 15 30 4 80 20 40 5 125 25 50 6 180 30 60 1 2 3
• Uniform acceleration For . Acceleration = rate of change of velocity a = = = = 10m.s -2 V v 2 – v 1 60 - 0 t t 2 - t 1 6 Time (s) Displacement (m) Average Velocity Instantaneous Velocity (m.s-1) Acceleration (m.s-2) 0 0 0 0 10 5 5 10 10 20 10 20 10 45 15 30 10 4 80 20 40 10 5 125 25 50 10 6 180 30 60 10 1 2 3