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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)
4
50km/h
0m 10m 20m 30m 40m 50m 60m 70m
www.rta.nsw.gov.au/roadsafety/speedandspeedcameras/speedingresearch.html
29 metresNo Impact
5
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
6
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
7
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
8
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
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.