Alevel Mechanic question

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.

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?

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

$v'$ 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?

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.