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    a particle of mass m is projected towards a point O with initial speed √5/3 m/s from a point P where [OP]=3 metres. the particle is repelled from O by a force of magnitude 4m/x³ where x is its distance from O.
    (i) show that the equation of motion is v dv=4x^-3 dx
    (ii) find how close the particle will get to O
    (iii) find its speed when it has travelled half the distance from P to this nearest point
    i do not know even how to start this question off could someone show me the step to step guide to answer the qustion
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    (Original post by markosheehan)
    a particle of mass m is projected towards a point O with initial speed √5/3 m/s from a point P where [OP]=3 metres. the particle is repelled from O by a force of magnitude 4m/x³ where x is its distance from O.
    (i) show that the equation of motion is v dv=4x^-3 dx
    (ii) find how close the particle will get to O
    (iii) find its speed when it has travelled half the distance from P to this nearest point
    i do not know even how to start this question off could someone show me the step to step guide to answer the qustion
    You know that the force is F = -4m/x^3 (directed where?)

    So you also know that F= ma, well - you should.

    So ma = 4m/x^3. Now cancel the m's.

    So a = 4/x^3. We're getting close.

    But you also know that a = dv/dt.

    Using the chain rule, you can re-write this as a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v\frac{dv}{dx} since dx/dt is just the rate of change of displacement with respect to time, otherwise known as velocity.

    So plug this into the equation. You should be able to do the rest by yourself. If you can't, then post up your own thoughts and attempt.
 
 
 
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