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    Suppose that a boat moves in a straight line with speed v through water with resistance proportional to the square of its speed v with initial speed v_0 and initial position x_0. Find its speed v(t) and position x(t) and conclude that the boat travels an infinite distance, i.e. x(t) tends to infinity as t tends to infinity.

    This is the question I have been given. Using the equation of motion: "s= 1/2(u+v)t" I have got the equations:

    v(t)=(2x_0)/t-v_0 and x(t)=1/2(v_0+v)t

    My equation for x(t) does tend to infinity as t tends to infinity, but I was thinking that I haven't factored in the water resistance at all. I was thinking of using one of the other equations of motion, i.e "v^2=u^2+2as" that factors in the acceleration and say that the acceleration was -{v_0}^2 at x_0, but I am unsure whether this is the correct approach to the question.

    The only other thing I considered was saying that the forces in the x positive direction were= v and the forces in the negative x direction were= v^2 but this seemed far too simple.

    Any help?!?!
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    (Original post by Happy2Guys1Hammer)
    Any help?!?!
    Acceleration is not constant, so suvat does not apply.

    You need to work the situation from scratch. Work out the acceleration and start integrating.

    Note: There is no force in the +ve x direction, only the one in the -ve x direction.
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    (Original post by ghostwalker)
    Acceleration is not constant, so suvat does not apply.

    You need to work the situation from scratch. Work out the acceleration and start integrating.

    Note: There is no force in the +ve x direction, only the one in the -ve x direction.
    Okay well if that's the case then I think this is a kind of mechanics that I have not encountered before. If I remember correctly you differentiate acceleration to get speed and then again to get distance but I'm not sure how I would go about getting the acceleration.
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    (Original post by Happy2Guys1Hammer)
    Okay well if that's the case then I think this is a kind of mechanics that I have not encountered before. If I remember correctly you differentiate acceleration to get speed and then again to get distance but I'm not sure how I would go about getting the acceleration.
    False. The rate of change of displacement is velocity, and the rate of change of velocity is acceleration. Thus, you integrate acceleration to get speed and then again to get distance.

    ...Does this question really have anything to do with vectors?
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    (Original post by Happy2Guys1Hammer)
    but I'm not sure how I would go about getting the acceleration.
    This is rather a basic mechanics question, even for A-level, let along uni, and your notes should cover it. I'd suggest getting yourself familiar with them, if you're not already.

    But to start you need to be using F=ma, which you should be familiar with.
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    (Original post by ghostwalker)
    This is rather a basic mechanics question, even for A-level, let along uni, and your notes should cover it. I'd suggest getting yourself familiar with them, if you're not already.

    But to start you need to be using F=ma, which you should be familiar with.
    To be fair if I had notes, I wouldn't be asking for help.

    Okay so I will be using a=F/m which would give me "a=(v_0-{v_0}^2)/m"? and then I just integrate this with respect to t?
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    (Original post by Happy2Guys1Hammer)
    Suppose that a boat moves in a straight line with speed v through water with resistance proportional to the square of its speed v with initial speed v_0 and initial position x_0. Find its speed v(t) and position x(t) and conclude that the boat travels an infinite distance, i.e. x(t) tends to infinity as t tends to infinity.
    Just rewrite what you have been given to form equations for the variables.

    e.g. if the resistance is proportional to the square of speed, this means that the rate of change of speed (i.e. acceleration)
    satisfies

    \frac{dv}{dt} = \lambda v^2
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    (Original post by Mark85)
    Just rewrite what you have been given to form equations for the variables.

    e.g. if the resistance is proportional to the square of speed, this means that the rate of change of speed (i.e. acceleration)
    satisfies

    \frac{dv}{dt} = \lambda v^2
    Ah okay so as speed is equal to the integration of acceleration, this would mean that  v={\lambda}v^2t+c and x=({\lambda}v^2t^2)/2 + ct + d . Where c and d are constants. So, where do V_0 and x_0 come in to things?
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    (Original post by Happy2Guys1Hammer)
    Ah okay so as speed is equal to the integration of acceleration, this would mean that  v={\lambda}v^2t+c and x=({\lambda}v^2t^2)/2 + ct + d . Where c and d are constants. So, where do V_0 and x_0 come in to things?
    Hang on, is that the same 'v' on both sides of the equation?

    You can't just integrate it like that if v is a function of t - you need to separate variables in the differential equation and then integrate!

    Which Mechanics module is this, and what differential equation work have you done before?
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    (Original post by Happy2Guys1Hammer)
    Ah okay so as speed is equal to the integration of acceleration, this would mean that  v={\lambda}v^2t+c and x=({\lambda}v^2t^2)/2 + ct + d . Where c and d are constants. So, where do V_0 and x_0 come in to things?
    You probably ought to go back and brush up on the basics like the definitions of speed, acceleration, integration, differentiation, basic seperable ODEs and proportionality before tackling this question. Otherwise, we might as well be talking to you in Russian for all you understand of anything. You will have to apply some interpretation too since at some point you will have to decide whether \lambda should be positive or negative for instance.

    Velocity is a function of time and acceleration is the rate of change of velocity with respect to time.

    If you like, rewrite what I wrote as

    \frac{dv(t)}{dt} = \lambda v^2(t)

    This is a seperable ODE. As is usual with indefinite integrals, the set of solutions will parameterised by a constant. As is also usual, specifying an initial value of v(t) i.e. a value v_0 = v(0) specifies a particular solution inside that set.
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    (Original post by Mark85)
    You probably ought to go back and brush up on the basics like the definitions of speed, acceleration, integration, differentiation, basic seperable ODEs and proportionality before tackling this question. Otherwise, we might as well be talking to you in Russian for all you understand of anything. You will have to apply some interpretation too since at some point you will have to decide whether \lambda should be positive or negative for instance.

    Velocity is a function of time and acceleration is the rate of change of velocity with respect to time.

    If you like, rewrite what I wrote as

    \frac{dv(t)}{dt} = \lambda v^2(t)

    This is a seperable ODE. As is usual with indefinite integrals, the set of solutions will parameterised by a constant. As is also usual, specifying an initial value of v(t) i.e. a value v_0 = v(0) specifies a particular solution inside that set.
    Okay I think I am starting to understand now. Here's what I have done:

    \int dv/dt = \int \lambda v^2 \Rightarrow \int 1/v^2 dv= \int \lambda dt \Rightarrow -1/v= \lambda t+c \Rightarrow v=-1/( \lambda t+c)

    Here is where I have hit another problem though as I set V_0=V(0) as you said, but that gives V=-1/c. How do I go about calculating the constant? .
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    You weren't given a numerical value for the initial value v_0.

    Therefore all you can do is express the constant of integration in terms of it.
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    (Original post by Mark85)
    You weren't given a numerical value for the initial value v_0.

    Therefore all you can do is express the constant of integration in terms of it.
    Okay thank you very much . I think I will be good from here on out.
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    (Original post by Mark85)
    You weren't given a numerical value for the initial value v_0.

    Therefore all you can do is express the constant of integration in terms of it.
    Okay I have one more question, I think this is kind of similar...
    A parachutist whose mass is 75kg drops from a helicopter hovering 4000m above the ground and falls towards the earth under the influence of gravity. Assume that the force due to air resistance is proportional to the speed of the parachutist with proportionality constant k1 = 15kg/s when the chute is closed and with constant k2 = 105kg/s when the chute is open. If the parachute does not open until 1 minute after the parachutist has left the helicopter, after how many seconds will the parachutist land on the ground?

    I would approach this question in a similar way, but due to the introduction of weight given to the parachutist. I would approach this using F=ma to get the acceleration equal to a=(kv/75) and use constant acceleration formulae from there to solve further and hence find t when v is equal to 0.
 
 
 

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