Question

A streetcar moves rectilinearly from station A (here car stop) to next station B (here also car stops) with an acceleration varying according to the law $ F=a-bx $ where $ a $ and $ b $ are possible constant and $ x $ is the distance from station A. If the maximum distance the two station is $ x=\dfrac{Na}{b} $ then final $ N. $

Answer 1

Hint :Acceleration is the rate of change of velocity acceleration means the speed is changing, but always when an object moves in circular both are constant speed. It is still accelerating, because the direction of its velocity is changing.
 $ \therefore $ The acceleration is $ F=a-bx $
 $ \therefore V=\dfrac{d}{t} $
By using acceleration formula,
 $ F=a-bx....(i) $
The acceleration is decreasing with increasing in distance $ x. $
Hence, the velocity will be maximum.
Where, $ F=0 $
From equation $ (i) $
 $ a=bx=0 $
 $ x=\dfrac{a}{b} $ .

Complete Step By Step Answer:
Given,
 $ F=a-bx,F=0 $ (velocity is maximum)
 $ x=\dfrac{a}{b} $
Now we know
 $ F=\dfrac{dv}{dt} $
 $ F=\dfrac{dv}{dt}\times \dfrac{dx}{dx} $
 $ F=V\dfrac{dv}{dx} $ $ \therefore F=a-bx $
 $ a-bx=\dfrac{Vdv}{dx} $
 $ Vdv=\left( a-bx \right)dx...(ii) $
Now for maximum velocity
 $ \int\limits_{0}^{{{V}_{\max }}}{Vdv=\int\limits_{0}^{\dfrac{a}{b}}{\left( a-bx \right)dx}} $
 $ {{\left[ \dfrac{{{V}^{2}}}{2} \right]}^{{{V}_{\max }}}}=\left( ax-b\dfrac{{{x}^{2}}}{2} \right)_{0}^{\dfrac{a}{b}} $
 $ \dfrac{V_{\max }^{2}}{2}=\dfrac{{{a}^{2}}}{b}-\dfrac{b{{a}^{2}}}{2{{b}^{2}}} $
 $ \dfrac{V_{\max }^{2}}{2}=\dfrac{2b{{a}^{2}}-b{{a}^{2}}}{2{{b}^{2}}} $
 $ \dfrac{V_{\max }^{2}}{2}=\dfrac{{{a}^{2}}}{2b} $
 $ V_{\max }^{2}=\dfrac{{{a}^{2}}}{b} $
 $ {{V}_{\max }}=\dfrac{a}{\sqrt{b}} $
Now, for the maximum distance $ x $ between the two station on integrating of equation (a)
At A and B, cars at race. So the velocity is zero at both stations.
Now,
 $ \int\limits_{0}^{0}{Vdv=\int\limits_{0}^{x}{\left( a-bx \right)dx}} $
 $ 0=\left( ax-b\dfrac{{{x}^{2}}}{2} \right)_{0}^{x} $
 $ ax-b\dfrac{{{x}^{2}}}{2}=0 $
 $ a-b\dfrac{x}{2}=0 $
 $ x=\dfrac{2a}{b}...(ii) $
And given that,
 $ x=\dfrac{Na}{b}...(iii) $
Now, on comparing equation $ (ii)\And (iii) $
Then,
 $ N=2 $
Hence, the value of $ N $ is $ 2. $

Additional Information:
All the equations of Newtonian metonics are based on Newton law of motion. All the equations are a Derivation and all theories are governed by them, therefore Newtonian Mechanics only considers the larger objects on which the gravitational field works.

Note :
The common differences between two stations $ A\And B $ is of acceleration, therefore the equation in the question can be used to determine velocity, acceleration or displacement of a body at any instant of time and then the value of is obtained by integrating the equation.