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Topic: Maximum Separation (Read 326 times) |
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navdeep1771
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Let your thoughts go beyond your imagination
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Maximum Separation
« on: Aug 12th, 2021, 10:11pm » |
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Two bodies move in a straight line towards each other at initial velocities v1 and v2 and with constant accelerations a1 and a2 directed against the corresponding velocities at the initial instant. What must be the maximum initial separation Lmax between the bodies for which they meet during the motion?
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rmsgrey
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Re: Maximum Separation
« Reply #1 on: Aug 13th, 2021, 7:29am » |
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Answer: Lmax=(v1 + v2)2/(2(a1 + a2)) Working: ::hidden: | Working in a frame where one of the bodies is stationary throughout (and ignoring relativistic corrections), the other body has initial velocity, u=v1+v2, and constant acceleration in the opposite direction, a=a1+a2. The change in separation between the two bodies, s, at a given time, t, is given by standard SUVAT formulae: s=ut-at2/2 Since s reaches a maximum when the closing velocity, v=u-at, is 0, we get: u-at=0 t=u/a Substituting for t in the equation for s gives: s=u(u/a) - a(u/a)2/2 =u2/a - u2/2a =u2/2a and substituting back for u and a gives the answer above: Lmax=(v1 + v2)2/(2(a1 + a2)) | :: edit: the subscript tags are behaving weirdly, and I can't get them to work properly edit2: fixed. Thanks towr.
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« Last Edit: Dec 29th, 2021, 11:22am by rmsgrey » |
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towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
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Re: Maximum Separation
« Reply #2 on: Dec 29th, 2021, 7:46am » |
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The tags are misbehaving because the board inserts spaces in a too-long string without any.
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