Question

A street car moves rectilinearly from station $A$ (here car stops) to next station $B$ (here also car stops) with an acceleration varying according to the law $f=a-bx$, where $a$ and $b$ are positive constants and $x$ is the distance from station $A$. If the maximum distance between the two stations is $x=\frac{Na}{b}$ then find $N$.

Answer 1

Given that,

The acceleration is $f=a-bx.....\left(I\right)$

The acceleration is decreasing with increasing x.

Hence, the velocity will be maximum

Where, f= 0,

From equation (I),

$a-bx=0$

$x=\frac{a}{b}$

Now, we know that

$f=\frac{dv}{dt}$

$f=\frac{dv}{dt}\times \frac{dx}{dx}$

$f=v\frac{dv}{dx}$

Now, we can write,

$v\frac{dv}{dx}=a-bx$

$vdv=(a-bx)dx.....\left(II\right)$

Now, for maximum velocity,

$\underset{0}{\overset{v_{max}}{\int }}vdv=\underset{0}{\overset{\frac{a}{b}}{\int }}(a-bx)dx$

${\left[\frac{v^2}{2}\right]}_0^{v_{max}}={(ax-b\frac{x^2}{2})}_0^{\frac{a}{b}}$

$\frac{v_{max}^2}{2}=\frac{a^2}{b}-\frac{b{a}^2}{2b^2}$

$\frac{v_{max}^2}{2}=\frac{2b{a}^2-b{a}^2}{2b^2}$

$\frac{v_{max}^2}{2}=\frac{a^2}{2b}$

$v_{max}^2=\frac{a^2}{b}$

$v_{max}=\frac{a}{\sqrt{b}}$

Now, for the maximum distances x between the two stations

On integrating of equation (II)

At A and B, cars at rest so the velocity is zero at the both stations.

Now,

$\underset{0}{\overset{0}{\int }}vdv=\underset{0}{\overset{x}{\int }}(a-bx)dx$

$0={(ax-b\frac{x^2}{2})}_0^x$

$ax-b\frac{x^2}{2}=0$

$a-b\frac{x}{2}=0$

$x=\frac{2a}{b}......\left(III\right)$

And given that,

$x=\frac{Na}{b}....\left(IV\right)$

Now, on comparing equation (III) and (IV)

Then,

$N=2$

Hence, the value of $N$ is $2$.