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Thread: Mechanics problem: ratios of velocities, times and distances

  1. #1
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    Question Mechanics problem: ratios of velocities, times and distances

    I am working through an old mechanics textbook for UK A-level (1970s/80s) and am having difficulty with following problem, which is from an old A-level paper. I have managed to solve the first part of the question, but am floundering on the second and third parts. Help please!

    This is the first part of the question:
    A particle moves on a straight line so that its distance s from a fixed point of the line is given by s = b sin kt, where t is the time and b, k are constants. Prove that its velocity v at any point is given by v2=k2(b2 - s2).

    and here is my answer:

    s = b sin kt so ds/dt = v = bk cos kt
    cos kt = v/bk, so sin2 kt = 1 - v2 / b2k2 = (b2k2 - v2) / b2k2

    Therefore

    s2/b2 = (b2k2 - v2) / b2k2 so k2s2 = b2k2 - v2 i.e. v2 = k2(b2 - s2)


    Here are the second and third parts of the question, where I need help:

    Particles A and B move along a straight line, each starting from a point O with velocity u. A is acted on by a constant force directed towards O, and B by a force, also towards O, proportional to the distance from O.

    (i) If they both cover the same distance L before starting back towards O, show that the times taken tA and tB respectively, are in the ratio 4 : pi, and that their velocities as they pass a point distant s from O are in a ratio given by

    vA2 : vB2 = L : (L + s)

    (ii) If they both take the same time T between leaving O and returning to O, show that their maximum distances from O are in a ratio given by

    LA : LB = pi : 4

    I have tried using F=ma and integrating the expressions for acceleration with respect to t and with respect to s, but I can't get to the ratios stated, particularly with the time expressions.

    Any help much appreciated.
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  2. #2
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    Lightbulb Re: Mechanics problem: ratios of velocities, times and distances

    OK, I've solved it. Particle A has constant deceleration away from O, so obeys the equations of motion for constant acceleration, and we get

    vA2 = u2(L - s)/L

    Particle B's motion can be found by integrating with respect to v and s, to get

    vB2 = u2(L2 - s2)/L2

    The ratio of these two is vA2 : vB2 = L : L + s

    For particle A, solving for time gives tA = 2L/u

    For particle B, noting the result of the first part of the question yields sB = L sin ut/L and solving for time gives tB = L(pi)/(2u)

    The ratio of these two times is tA : tB = 4 : pi
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