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

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

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

2 2 End of Slide 50 km/h 23 m 60 km/h 30 m 80 km/h 47 m 100 km/h 70 m

3 3 www.liikenneturva.fi/www/en/animaatiot/pysahtymismatka_ENG.swf

4 4 50 km/h 0m10m20m30m40m50m60m70m www.rta.nsw.gov.au/roadsafety/speedandspeedcameras/speedingresearch.html 29 metres No Impact

5 5 60 km/h 0m10m20m30m40m50m60m70m www.rta.nsw.gov.au/roadsafety/speedandspeedcameras/speedingresearch.html 38 metres Impact at 44 km/h

6 6 70 km/h 0m10m20m30m40m50m60m70m www.rta.nsw.gov.au/roadsafety/speedandspeedcameras/speedingresearch.html 48 metres Impact at 62 km/h

7 7 80 km/h 0m10m20m30m40m50m60m70m www.rta.nsw.gov.au/roadsafety/speedandspeedcameras/speedingresearch.html 59 metres Impact at 78 km/h

8 8 90 km/h 0m10m20m30m40m50m60m70m www.rta.nsw.gov.au/roadsafety/speedandspeedcameras/speedingresearch.html 70 metres Impact at 90 km/h End of Slide

9 9 End of Slide http://www.auburn.nsw.gov.au/uploadedFiles/Stopping%20Distance%20v08%281%29.pdf

10 10 End of Slide http://www.auburn.nsw.gov.au/uploadedFiles/Stopping%20Distance%20v08%281%29.pdf

11 11 End of Slide

12 12 End of Slide A car travelling at 100 km/h [28 m/s] breaks with a deceleration of 14m/s 2. How long does it take to stop after breaking? v = u + at Final Velocity 0 m/s Starting Velocity 100 km/h Deceleration 5 m/s 2 Time ?

13 13 End of Slide How long does it take to stop after breaking? v = u + at v = 0 m/s u = 28 m/s a = -14 m/s 2 0 =28- 14t +14t 14t = 28 ÷14 t = 2 seconds A car travelling at 100 km/h [28 m/s] breaks with a deceleration of 14m/s 2.

14 14 End of Slide How far does it take to stop after breaking? s = ut + ½at 2 v = 0 m/s u = 28 m/s a =-14 m/s 2 s = t = 2 s 28x 2x 2+ 0.5x -14x 22x 22 = 28 metres A car travelling at 100 km/h [28 m/s] breaks with a deceleration of 14m/s 2. Distance

15 15 End of Slide What is the total stopping distance? v = 0 m/s u = 28 m/s a = -14 m/s 2 t = 2 s s = 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/s 2.

16 16 End of Show Drive a vehicle with ABS. [Anti-locking Breaking System] Video


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