1 press ctrl-a ©g dear 2011 – not to be sold/free to use stopping distances stage 6 - year 11...

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1 Press Ctrl-A Press Ctrl-A ©G Dear 2011 – Not to be sold/Free to use Stopping Distances Stopping Distances Stage 6 - Year 11 Stage 6 - Year 11 Applied Mathematic (Preliminary General 1)

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Press Ctrl-APress Ctrl-A©G Dear 2011 – Not to be sold/Free to use

Stopping DistancesStopping DistancesStage 6 - Year 11Stage 6 - Year 11

Applied Mathematic(Preliminary General 1)

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End ofSlide

50 km/h23 m

60 km/h30 m

80 km/h47 m

100 km/h70 m

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www.liikenneturva.fi/www/en/animaatiot/pysahtymismatka_ENG.swf

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50km/h

0m 10m 20m 30m 40m 50m 60m 70m

www.rta.nsw.gov.au/roadsafety/speedandspeedcameras/speedingresearch.html

29 metresNo Impact

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60km/h

0m 10m 20m 30m 40m 50m 60m 70m

www.rta.nsw.gov.au/roadsafety/speedandspeedcameras/speedingresearch.html

38 metresImpact at 44 km/h

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70km/h

0m 10m 20m 30m 40m 50m 60m 70m

www.rta.nsw.gov.au/roadsafety/speedandspeedcameras/speedingresearch.html

48 metresImpact at 62 km/h

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80km/h

0m 10m 20m 30m 40m 50m 60m 70m

www.rta.nsw.gov.au/roadsafety/speedandspeedcameras/speedingresearch.html

59 metresImpact at 78 km/h

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90km/h

0m 10m 20m 30m 40m 50m 60m 70m

www.rta.nsw.gov.au/roadsafety/speedandspeedcameras/speedingresearch.html

70 metresImpact at 90 km/h

End ofSlide

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End ofSlide

http://www.auburn.nsw.gov.au/uploadedFiles/Stopping%20Distance%20v08%281%29.pdf

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End ofSlide

http://www.auburn.nsw.gov.au/uploadedFiles/Stopping%20Distance%20v08%281%29.pdf

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End ofSlide

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End ofSlide

A car travelling at 100 km/h [28 m/s] breaks with a deceleration of 14m/s2.

How long does it take to stop after breaking?

v = u + at

FinalVelocity0 m/s

StartingVelocity

100 km/h

Deceleration5 m/s2

Time ?

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End ofSlide

How long does it take to stop after breaking?v = u + at

v = 0 m/su = 28 m/sa = -14 m/s2

0 = 28 - 14t+14t +14t

14t = 28÷14 ÷14

t = 2 seconds

A car travelling at 100 km/h [28 m/s] breaks with a deceleration of 14m/s2.

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End ofSlide

How far does it take to stop after breaking?

s = ut + ½at2

v = 0 m/su = 28 m/sa =-14 m/s2

s =

t = 2 s

28x 2 + 0.5x -14x 22

= 28 metres

A car travelling at 100 km/h [28 m/s] breaks with a deceleration of 14m/s2.

Distance

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End ofSlide

What is the total stopping distance?

v = 0 m/su = 28 m/sa = -14 m/s2

t = 2 ss = 28 m

Reaction Time = 1.2 seconds

Distance = 1.2sx 28 m/s

= 34 metres

Total = 34 +28

= 62 metres

A car travelling at 100 km/h [28 m/s] breaks with a deceleration of 14m/s2.

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End ofShow

Drive a vehicle with ABS.[Anti-locking Breaking System]

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