Question

A street car moves rectilinearly from station A (here car stops) to the 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. The distance between the two stations & the maximum velocity are respectively.

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

Answer 1

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

$f=a-bx$ means particle will do SHM.
For maximum velocity, acceleration should be zero.
i.e. $a-bx=0\Rightarrow x=\frac{a}{b}$
$\therefore$ At $x=\frac{a}{b}$, the particle has its maximum velocity.
$f=\frac{vdv}{dx}=a-bx$
$f=\int vdv=\int (a-bx)dx$
$\Rightarrow \frac{v^2}{2}=ax-\frac{b{x}^2}{2}+cAtx=0;v=0\Rightarrow c=0$
Substituting $x=\frac{a}{b}$ gives
$v_{max}=\frac{a}{\sqrt{b}}$
Also, the velocity of the car should become zero at station B.
i.e $ax-\frac{b{x}^2}{2}=0\Rightarrow x=0;x=\left(\frac{2a}{b}\right)$
$\therefore$ Distance between the two stations is $\frac{2a}{b}$


Alternate :
$f=a-bx$ means particle will do SHM.
At mean position ;
$f=0$
$\Rightarrow x=\frac{a}{b}$

In the figure shown, 'C' is the mean position and A & B are extreme positions
$\therefore x_{max}=\frac{2a}{b}\&V_{max}=\omega A=\sqrt{b}.\frac{a}{b}=\frac{a}{\sqrt{b}}$