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B Finding velocity in terms of displacement

  1. Jun 2, 2016 #1
    I know two methods as to how to do this, but one of them doesn't seem to work.
    Say you're given x = 3t^2 + 6t - 4.
    If you treat x as an unknown variable and movie it over to the other side, you can use completing the square to solve for t. This eventually gets you: t = 1/3*sqrt(3(x+7)) - 1
    Then you differentiate the initial function to obtain the velocity, substitute the calculated time in and thus obtain velocity in terms of displacement. This method ultimately got me v = 2*sqrt(3x+21), which works and is completely fine.

    However, when I try the alternative way it doesn't work. If we go back to the beginning and find both velocity and acceleration, we get x = 3t^2 + 6t - 4, v = 6t + 6 and a = 6. To obtain dv/dx, we need to multiple the dv/dt by dt/dx (with the dt on the numerator and denominator canceling out). We know dv/dt, as that's just another way of expressing the acceleration, which is 6. Now this is the part which I think is messing up; I contend that dt/dx would should just be the inverse of dx/dt, which we know to be 6t + 6. This should mean that dt/dx should just equal 1/(6t + 6). Thus, dv/dx = dv/dt * dt/dx = 6*(1/(6t + 6)) = 1/(t+1). To obtain velocity in terms of displacement, we then would need to differentiate dv/dx, which should leave us with v(x) = ln(|t + 1|) + c. At t = 0, we know velocity = 6. Therefore, 6 = ln(|1|) + c, c = 6. v(x) = ln(|t+1|) + 6.

    However, this equation is clearly very different from the one obtained in the first method and doesn't work. Can someone please tell me how to fix the second method?
  2. jcsd
  3. Jun 3, 2016 #2
    I think you are given an equation for x as a function of time t ..... however if you plot x as function of t you have initial condition for the motion described.
    at t=0 the position of the body is at x= -4 and at t=t1 and t2 ,two values the body is at x=0 i.e. the origin as you have a quadratic in t.

    one of them is real...
    moreover if you place the origin at -4 the acceleration is 6 units and and the initial velocity is also 6 units.
    so you have full description of the motion.
    you are doing some exercises to again describe the motion .pl.check if you get a different result whether your initial condition is changed !
  4. Jun 3, 2016 #3


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    It is possible that ##\frac{dt}{dx}## is the derivative of ##t(x)=\frac{1}{3}\sqrt{3(x+7)}-1## respect ##x##, multiply this quantity by ##6## and look what happen ...
  5. Jun 3, 2016 #4
    i think here you are making an error...
    you have dv/dx = a function of t and now you are integrating dv on the left and dx . function of t on the right so you must first convert function of t in terms of x and then do the integration ...and i hope you will get consistent behaviour of velocity as function of x.
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