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Math Help - Motion with uniform acceleration??

  1. #1
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    Unhappy Motion with uniform acceleration??

    Hi, my first post dont know if im on the right one :S.

    Can someone help with this maths question, currently using SUVAT method.

    A train leaves a station from rest with a constant acceleration of 0.3ms-1. It reaches its maximum speed after 90 seconds and mantains this speed for a futher 5 minutes, when it slows down to a stop at a second station with an acceleration of -1ms-2. How far apart are the two station.

    Ive figured that the first 90 seconds its travelled 1215m I think. Can anyone help me with the rest. If you could to explain how you did it. I want to understand. Thankyou.
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  2. #2
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by NathanBUK View Post
    Hi, my first post dont know if im on the right one :S.

    Can someone help with this maths question, currently using SUVAT method.

    A train leaves a station from rest with a constant acceleration of 0.3ms-1. It reaches its maximum speed after 90 seconds and mantains this speed for a futher 5 minutes, when it slows down to a stop at a second station with an acceleration of -1ms-2. How far apart are the two station.

    Ive figured that the first 90 seconds its travelled 1215m I think. Can anyone help me with the rest. If you could to explain how you did it. I want to understand. Thankyou.
    By the "SUVAT method" I presume that's some acronym that somehow encapsulates v = u + at, s = ut+\frac 1 2 a t^2, v^2 = u^2 + 2 a s.

    Split the thing up into 3 bits. I'm leaving out the units (metres, seconds, etc.) for clarity, but be aware your teacher will crucify you if you do the same.

    1. You have a = 0.3, u = 0, t = 90. From that you can easily calculate the distance by s = ut + \frac 1 2 a t^2 (which you did, whether you got the right answer depends on how accurately you punched your calculator keys). Now, what you *also* want to do is get the velocity at this point - so use v = u + at for that (and you know u = 0 and a = what I said above.

    2. Now you know how fast it's going, and you've got a = 0 so s = ut (+ \frac 1 2 a t^2) and that gives you the second distance.

    3. In this case, you have v^2 = u^2 + 2 a s again, but this time a = -1, u is the velocity from the last bit, and v = 0. Then rearrange the equation so it's in terms of s.

    Add them up together and you're done.
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  3. #3
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    Oh Thankyou soo much!! Just starting out on this and well homework and no teacher come a bit stuck but thankyou so much!!
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