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I am trying to model a situation in which a car passes the point A on a straight road with a speed of 10 m/s, and moves with a constant acceleration of a m/s^2 along the road for T seconds until it reaches the point B where its speed is V m/s. The car is then travelling at this speed (V) for a further 10 seconds when it reaches a point C. From C it travels for a further T seconds with a constant acceleration 3a m/s^2, until a speed of 20 m/s is reached at a point D. The question being asked is to show that V = 12.5 m/s. However, I feel as if I am doing the graph wrong. Here is what my graph looks like:

enter image description here

Is this graph ok, and if so what do I need to do in order to find V, because I have been trying to use three SUVAT equations as there are three variables (which means I need three equation to solve simultaneously) to get V, a and T, however, I haven't really made any progress. Is there a way to find a solution using the graph?

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  • $\begingroup$ Yes the question is to find V $\endgroup$ – Benny Jul 8 '18 at 4:30
  • $\begingroup$ Th distance between A and D is given in the second part of the question, it is 675m $\endgroup$ – Benny Jul 8 '18 at 4:35
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enter image description here $$\triangle V =at+3at=20-10=10\mbox{ m/s}$$ $$\mbox{So, }at=2.5 \mbox{ m/s}$$

$$V=10+at=12.5\mbox{ m/s}$$ $$675=10t+\dfrac12at^2+(12.5)(10)+12.5t+\left(\dfrac12\right)3at^2$$ Now plug in $at=2.5$ $$675=10t+\frac12(2.5t)+125+12.5t+\frac32(2.5t)$$ Now solve for $t$

Can you take it from here?

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  • $\begingroup$ Yes, I understand how you have done it with the distance from A to D, however, as this information is given the second part of the question is it possible to work out V without it? $\endgroup$ – Benny Jul 8 '18 at 4:47
  • $\begingroup$ Oh I see, cheers, but I don't quite get where 3at comes from $\endgroup$ – Benny Jul 8 '18 at 4:59
  • $\begingroup$ $\triangle V =at+3at$ $\endgroup$ – Key Flex Jul 8 '18 at 5:16

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