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Applied Maths LC Paper 1991 Q1 Linear Motion HELP!!!!

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  • 07-01-2015 10:48pm
    #1
    leavingcerttt
    Registered Users Posts: 2 ✭


    hi could someone please explain how to do this question?

    1991
    (a) A particle starts from rest at a point p and accelerates at 2m/s2 until it reaches a speed v m/s. It travels at this speed for 1 minute before decelerating at 1 m/s2 to rest at q. The total time for the journey is 2 minutes.
    (i) Calculate the distance pq. (ans=3600m)
    (ii) If a second particle starts from p at time t = 0 and moves along pq with speed (2t + 50)m/s, find the time taken to reach q. (ans=40secs)

    THANK YOU!
    Tagged:


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    #2
    Raspberry Fileds
    Closed Accounts Posts: 294 ✭✭Raspberry Fileds


    Struggling with both parts?


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    #3
    Raspberry Fileds
    Closed Accounts Posts: 294 ✭✭Raspberry Fileds


    App_Maths.png
    t1 + t3 = 60
    a = acceleration = 2
    b = deceleration = 1

    Tan A = a = v/t1
    Tab B = b = v/t3

    v/2 + v/1 = t1 + t3
    3v/2 = 60
    v = 40

    1/2(t1)(v) + 1/2(t3)(v) + (t2)(v) = d
    1/2(t1+t3)(v) + (t2)(v) = d
    1/2(60)(40) + (60)(40) = d = 3600


    ii) The mistake many will have made is assuming the second particle started from rest.

    At t=0, v = 2(0) + 50 = 50.

    The coefficient of t gives the acceleration - i.e. 2.

    u = 50
    v = 2t + 50
    a = 2
    s = 3600
    t = t

    3600 = 50t + 1/2(2)(t^2)
    (t^2) + 50t - 3600 = 0

    The quadratic formula will give you two roots - one of which will be negative. As time can only increase, and the particle reaches q after t=0, the positive root is the answer.


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