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mechanics help

  • 16-08-2007 3:39pm
    #1
    JoseJones
    Registered Users, Registered Users 2 Posts: 612 ✭✭✭


    can anyone help me with this question? I'm trying to study for my repeat exam and it's doing my head in...


Comments

  • #2
    Michael Collins
    Moderators, Science, Health & Environment Moderators Posts: 1,851 Mod ✭✭✭✭Michael Collins


    For the first part you just need to find the tangent to that curve at the point in question (given on your graph) i.e. you need to differentiate that curve, and find the slope at that point, the actual angle you get with the x-axis is then just the inverse trig tangent of that slope.

    The first bit of the second part is a trick question from what I can see. But sure have a go at the first part anyway and post up what you get...


  • #3
    JoseJones
    Registered Users, Registered Users 2 Posts: 612 ✭✭✭JoseJones


    ok, so i got dy/dx=0.1x, where x=10 => m=1 => 45deg....is this right? I'm always wary when I get 45deg as an answer, surely I've done something wrong? for the second part, it says in the question, a=2m/s^2....is that it? so how would I find the direction? what I did was split a into n-t co-ordinates. because the angle is 45, a(t) and a(n) are equal, so the direction of a is vertically down at 90deg....I'm probably way off tho....


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