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    So the question is as follows; "A displacement, s metres, at time t seconds, of a particle from a fixed point is given by s=2t^2+3t-2.
    i) Find the initial displacement and initial velocity.
    ii) Find any times where the velocity is zero.
    iii) Find any times where the particle is at the origin."
    I have tried all 3 sections but for the first section i had a distance of -2 metres, for the second section I had a time of -0.75 seconds and for the third section I had 2 times of 0.5 seconds and -2 seconds. If anyone could help then that would be great as I'm not the strongest in Mechanics!
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    (Original post by abbiechantellex)
    So the question is as follows; "A displacement, s metres, at time t seconds, of a particle from a fixed point is given by s=2t^2+3t-2.
    i) Find the initial displacement and initial velocity.
    ii) Find any times where the velocity is zero.
    iii) Find any times where the particle is at the origin."
    I have tried all 3 sections but for the first section i had a distance of -2 metres, for the second section I had a time of -0.75 seconds and for the third section I had 2 times of 0.5 seconds and -2 seconds. If anyone could help then that would be great as I'm not the strongest in Mechanics!
    The answers you have calculated are all correct, well done!
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    (Original post by HapaxOromenon3)
    The answers you have calculated are all correct, well done!
    Wow, really?! They were negative and so I got very confused, is there any particular reason as to why they are negative?
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    (Original post by abbiechantellex)
    Wow, really?! They were negative and so I got very confused, is there any particular reason as to why they are negative?
    A negative displacement means it goes in the opposite direction to nornal. E.g. in coordinate geometry, (3,0) is 3 units to the right of the origin and (-3,0) is 3 units to the left of the origin.

    Since t=0 is the time when we started measuring values of s, a negative time means that if we assume the equation for s was valid even before we started measuring, then the event happened however many seconds before we started measuring. That is, t=-0.75 means that the velocity was zero 0.75 seconds before we (the experimenter) started measuring s, assuming that the particle's motion followed the same equation for s in terms of t even though this was before we had started the experiment. Hopefully that makes sense!
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    (Original post by HapaxOromenon3)
    A negative displacement means it goes in the opposite direction to nornal. E.g. in coordinate geometry, (3,0) is 3 units to the right of the origin and (-3,0) is 3 units to the left of the origin.

    Since t=0 is the time when we started measuring values of s, a negative time means that if we assume the equation for s was valid even before we started measuring, then the event happened however many seconds before we started measuring. That is, t=-0.75 means that the velocity was zero 0.75 seconds before we (the experimenter) started measuring s, assuming that the particle's motion followed the same equation for s in terms of t even though this was before we had started the experiment. Hopefully that makes sense!
    Yes that makes a lot more sense now, thank you so much!!
 
 
 
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