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    • Thread Starter

    So i am really stuck with these two questions:

    6) A particle P is moving along the x-axis with velocity v=4t-2t^2. When t=0, P is at x=3. Find a) the position of P when t=2 b)the maximum velocity attainted by P c) the distance OP when the velocity is maximum

    7)A particle P os moving along the x-axis. When t=0, the velocity of P is 4.5 m/s. At time ts the acceleration of P is given by (3t-6) m/s^2. Find the time when the particle returns to its starting point.

    For the 6th question, i did the first two parts but i think my method is wrong so will really appreciate it if someone does it in detail.
    • PS Helper

    PS Helper
    6a) You should know that x = \displaystyle \int v\ \mbox{d}t, and you can find the constant of integration given the initial conditions (so sub in x=3 and t=0 to find the constant of integration, and then t=2 to find the value of x)
    6b) You can find the maximum of a curve by differentiating... so do that
    6c) You can use the answers to part a and b to answer this

    7) Again you need to integrate and use the initial conditions to find something in the form x = f(t), and then sub in x=0 and solve to find the time.

    Start with the definition of velocity:

    v= dx/dt

    To get position from this, need to integrate with respect to time, i.e

    x = integral(4t-2t^2)dt

    Gives us

    x = 2t^2-2/3t^3 + constant

    the info in question says t=0, x = 3 so can see that constant = 3, giving

    x = 2t^2-2/3t^3 + 3

    as equation for position. From there it is putting in numbers for part a.

    for part b, remember that a maximum (or minimum) is found by setting the differential of something to 0. In this case dv/dt = 0 would be a good bet!
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