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    a road vehicle of mass M is initially at rest and then moves off in a straight line on a level road. The driving force is given by MF(1-e^(-at)),where t is time and F and a are positive constants. The road resistance is given by Mbv, where b is a posi constant and v is the vehicle speed.
    verify that v=F/b{1-(1/(a-b))(ae^(-bt)-be^(-at)}

    Any help will be appreciated.
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    so the acceleration is a, which can be written as dv/dt = Net force/Mass

    you cannot use SUVAT as the acceleration is varying.

    Net force = MF(1-e(-at)) - Mbv

    a = { MF(1-e(-at)) - Mbv } / M

    then you need to integrate a wrt t to find v
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    (Original post by george1999poi)
    a road vehicle of mass M is initially at rest and then moves off in a straight line on a level road. The driving force is given by MF(1-e^(-at)),where t is time and F and a are positive constants. The road resistance is given by Mbv, where b is a posi constant and v is the vehicle speed.
    verify that v=F/b{1-(1/(a-b))(ae^(-bt)-be^(-at)}

    Any help will be appreciated.
    You know that F_n=Ma=M\cdot \frac{dv}{dt}=MF(1-e^{-at})-Mbv (where F_n is the net force) hence you can deduce \frac{dv}{dt}.

    You want to verify that the expression for v is true by finding showing that the equality v'=F(1-e^{-at})-bv is true.
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    (Original post by RDKGames)
    You know that F_n=Ma=M\cdot \frac{dv}{dt}=MF(1-e^{-at})-Mbv (where F_n is the net force) hence you can deduce \frac{dv}{dt}.

    You want to verify that the expression for v is true by finding showing that the equality v'=F(1-e^{-at})-bv is true.
    That is a good idea. Let me have a try.
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    then you need to integrate a wrt t to find v[/QUOTE]

    Thank you for your reply.The only thing I am struggling with is that I don't know how to integrate the function since the right hand side is expressed as both velocity and time.Do u have any idea?
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    (Original post by RDKGames)
    You know that F_n=Ma=M\cdot \frac{dv}{dt}=MF(1-e^{-at})-Mbv (where F_n is the net force) hence you can deduce \frac{dv}{dt}.

    You want to verify that the expression for v is true by finding showing that the equality v'=F(1-e^{-at})-bv is true.
    It doesn't look the same. Here is what I got:dv/dtFae^(-bt))/(a-b)-aFe^(-at)/(a-b) which is not the same as the one you state?
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    (Original post by george1999poi)
    It doesn't look the same. Here is what I got:dv/dtFae^(-bt))/(a-b)-aFe^(-at)/(a-b) which is not the same as the one you state?
    v' looks correct. For the sake of keeping things as simple as possible, please don't expand out. Just leave it as \displaystyle v'=F\cdot \frac{a}{a-b}(e^{-bt}-e^{-at})

    Next, consider the RHS, the F(1-e^{-at})-bv, you want to simplify this down to get v' above.
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    The main issue is the v in the expression
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    (Original post by the bear)
    so the acceleration is a, which can be written as dv/dt = Net force/Mass

    you cannot use SUVAT as the acceleration is varying.

    Net force = MF(1-e(-at)) - Mbv

    a = { MF(1-e(-at)) - Mbv } / M

    then you need to integrate a wrt t to find v
    Not the approach I'd recommend to OP. He'd have displacement, time, and velocity in the expression, and we don't know anything useful about the displacement to help us with the question.
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    (Original post by RDKGames)
    v' looks correct. For the sake of keeping things as simple as possible, please don't expand out. Just leave it as \displaystyle v'=F\cdot \frac{a}{a-b}(e^{-bt}-e^{-at})

    Next, consider the RHS, the F(1-e^{-at})-bv, you want to simplify this down to get v' above.
    But we don't know the v? We want to find v in term of t?
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    (Original post by george1999poi)
    But we don't know the v? We want to find v in term of t?
    Yes we do know v, it's given right there. The whole 'verifying' part is using it to show the results are consistent as we expect them to be.
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    (Original post by RDKGames)
    Yes we do know v, it's given right there. The whole 'verifying' part is using it to show the results are consistent as we expect them to be.
    OMG I am so damn! Thank you!
 
 
 
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