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Math Help - differential equations

  1. #1
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    differential equations

    1. A particle, of mass 0.4kg, is projected vertically upwards with an initial speed of 30 m/s. at time ts after projection, the speed of the particle is v m/s
    and the magnitude of the air resistance acting on the particle is 0.2vN.

    a) Show that, while the particle is ascending, v satisfies the differential equation

    \frac{dv}{dt}=-9.8-0.5v

    b) determine v in terms of t



    Right so i'm a little new to differential equations so i'm unsure exactly what to do. I know that dv/dt is acceleration. I know that acceleration when the particle goes upwards verticle the force goes against gravity so thats where the -9.8 comes from. just not sure about the -0.5v


    for part b i will take an educated guess that you have to integrate that equation, however its worth 6 marks so i think maybe theres more to it?
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  2. #2
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    Quote Originally Posted by djmccabie View Post
    1. A particle, of mass 0.4kg, is projected vertically upwards with an initial speed of 30 m/s. at time ts after projection, the speed of the particle is v m/s
    and the magnitude of the air resistance acting on the particle is 0.2vN.

    a) Show that, while the particle is ascending, v satisfies the differential equation

    \frac{dv}{dt}=-9.8-0.5v

    b) determine v in terms of t
    (a) while traveling upward, net force on the particle ...

    ma = -(mg + 0.2v)

    divide every term by m ...

    a = -\left(g + \frac{0.2v}{m}\right)

    m = 0.4 kg ... g = 9.8 m/s^2

    \frac{dv}{dt} = -(9.8 + 0.5v)


    yes, integration is required for part (b) ...

    \frac{dv}{dt} = -(9.8 + 0.5v)

    \frac{dv}{dt} = -0.5(19.6 + v)

    separate variables ...

    \frac{dv}{19.6 + v} = -0.5 \, dt

    integrate both sides of the equation, use your initial condition to find the constant of integration, then finalize velocity as a function of time.
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