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Math Help - kinematics problem

  1. #1
    Senior Member furor celtica's Avatar
    Joined
    May 2009
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    kinematics problem

    Ok I need some guidance with the second part of this problem. I attached two graphs hastily constructed with paint and which are somewhat lopsided, so keep in mind that they are by no means correct geometrically.

    A train travels from a station P to the next station Q, arriving at Q exactly 5 minutes after leaving P. The (t,v) graph for the trainís journey is approximated by three straight line segments as shown in figure 1.
    Write down the acceleration of the train during the first minute of the journey (I found 0.5 ms^-2 which is correct)
    Find the distance from P to Q (I found 7200m which is correct)

    On one occasion when the track is being repaired, the train is restricted to a maximum speed of 10ms^-1 for the 2000m of track lying midway between P and Q. The train always accelerates and decelerates at the rate shown in figure 1. When not accelerating or decelerating or moving at the restricted speed of 10ms^-1, the train travels at 30ms^-1. Sketch the (t,v) graph for the journey from P to Q when the speed restriction is in force, and hence find how long the train takes to travel from P to Q on this occasion.

    I sketched figure 2 for this part. A=E=0.5x60x30=900m, C=2000m
    So B=D=(7200 Ė 1800 -2000)/2 = 1700
    From this I calculate the time taken over B: 1700= 0.5 (10 + 30) t and get t= 85s
    85x2 + 200 + 60 x 2 = 490s
    But this is wrong, Iím supposed to get 460! What is wrong?
    Attached Thumbnails Attached Thumbnails kinematics problem-figure-1.bmp   kinematics problem-figure-2.bmp  
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    The mistake you have made is to assume that the train must start to decelerate as soon as it reaches its maximum speed. In fact, the train takes 40s to decelerate from 30ms^-1 to 10ms^-1, during which time it travels 800m. So during the acceleration and deceleration phases the train travels 900 + 800 = 1700 metres. But it does not have to slow down to 10ms^-1 until it has travelled 2600m. So it is able to go at maximum speed 30ms^-1 for a distance of 900m. The graph of speed against time should look like this:

    \setlength{\unitlength}{1.2mm}<br />
\begin{picture}(60,40)<br />
\put(0,5){\line(1,0){60}}<br />
\put(5,0){\line(0,1){40}}<br />
\thicklines<br />
\put(5,5){\line(1,5){6}}<br />
\put(11,35){\line(1,0){3}}<br />
\put(14,35){\line(1,-5){4}}<br />
\put(18,15){\line(1,0){20}}<br />
\put(38,15){\line(1,5){4}}<br />
\put(42,35){\line(1,0){3}}<br />
\put(45,35){\line(1,-5){6}}<br />
\put(3,3){$\scriptstyle 0$}<br />
\put(1,14){$\scriptstyle 10$}<br />
\put(1,34){$\scriptstyle 30$}<br />
\end{picture}
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