Question

A streetcar moves rectilinearly from station $A$ to the next station $B$ with an acceleration varying according to the law $f=a-bx$, where $a$ and $b$ are constant, and $x$ is the distance from station $A$. The distance between the two stations and maximum velocity are

• A. $x=\frac{2a}{b},v_{max}=\frac{a}{\sqrt{b}}$
• B. $x=\frac{b}{2a},v_{max}=\frac{a}{b}$
• C. $x=\frac{a}{2b},v_{max}=\frac{b}{\sqrt{a}}$
• D. $x=\frac{a}{b},v_{max}=\frac{\sqrt{a}}{b}$

Answer 1

The correct option is A $x=\frac{2a}{b},v_{max}=\frac{a}{\sqrt{b}}$

Acceleration of car, $f=a-bx$
Since acceleration, $f=v\frac{dv}{dx}$
$\Rightarrow v\frac{dv}{dx}=a-bx$
or ${\int }^{}vdv={\int }^{}(a-bx)dx$
or $\begin{array}{}\frac{v^2}{2}=ax-\frac{b{x}^2}{2}& \text{(1)}\end{array}$
$\Rightarrow v=\sqrt{2ax-b{x}^2}$
For $v$ to be maximum, $\frac{dv}{dx}=0$
$\Rightarrow \frac{d}{dx}\left(\sqrt{2ax-b{x}^2}\right)=\frac{2a-2bx}{2\sqrt{2ax-b{x}^2}}=0$
$\Rightarrow 2a-2bx=0$
$x=\frac{a}{b}$
Substituting the $x=\frac{a}{b}$ in equation (1)
$\frac{v_{max}^2}{2}=a\frac{a}{b}-\frac{b{\left(\frac{a}{b}\right)}^2}{2}=\frac{a^2}{2b}$
$v_{\text{max}}=\frac{a}{\sqrt{b}}$
Car will be at rest when velocity is zero, $v=0$.
Using equation $\left(1\right)$, we get
$ax-\frac{b{x}^2}{2}=0$
$x=\frac{2a}{b}$
Distance between two stations $=\frac{2a}{b}$