Question

A street car moves rectilinear from station $A$ to the next station $B($from rest to rest $)$ with an acceleration varying according to the law $f=a-bx;$ where $a$ and $b$ are constants and $x$ is the distance from station $A$. The distance between the two stations and the maximum velocity are

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

$f=a-bx$
We know that Acceleration, $f=\frac{dv}{dt}=\frac{dv}{dx}\times \frac{dx}{dt}=v\frac{dv}{dx}$ by using chain rule.
Thus, $v\frac{dv}{dx}=a-bx$
Integrate to get,
$\frac{v^2}{2}=ax-\frac{b{x}^2}{2}+C$
Now, initially$,x=0$ and $v=0$. Thus$,C=0$.
Now, at the second time when $v=0,$ we have $0=ax-\frac{b{x}^2}{2}$ or, $x=\frac{2a}{b}$. This is the total distance from $A$ to $B,$ because $v=0$ when it reaches $B$.
Also, maximum velocity means $f=0,$ thus $x=a/b$. For this x, we have $\frac{v^2}{2}=a\left(a/b\right)-\frac{b{\left(a/b\right)}^2}{2}=\frac{a^2}{b}-\frac{a^2}{2b}=\frac{a^2}{2b}$
or, $v=\frac{a}{\sqrt{b}}$