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 $f=a-b x$, 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:

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

Answer 1

Answer: (A)

Given that acceleration of car is
$f=a-b x$
For maximum velocity, acceleration should be zero.
$\Rightarrow a-b x=0$
$\Rightarrow x =\frac{a}{b}$
At $x=\frac{a}{b}$, the particle has its maximum velocity.
We use $f=\frac{v d v}{d x}=a-b x$
$\Rightarrow \frac{v^{2}}{2}=a x-\frac{b x^{2}}{2}+c$
At $x=0 ; v=0$
$\Rightarrow c=0$
Substituting $x=\frac{a}{b}$ gives maximum velocity as
$v_{\max }=\frac{a}{\sqrt{b}}$
Also, the velocity of the car should become zero at station $B$.
$\Rightarrow a x-\frac{b x^{2}}{2}=0$
$\Rightarrow x=0 ; x=\left(\frac{2 a}{b}\right)$
$\Rightarrow$ Distance between the stations is $\frac{2 a}{b}$.
Alternate :$f= a-bx$ means particle will execute SHM.
At mean position;$f= 0$

$\Rightarrow x=\frac{a}{b}$
In the figure shown, ${ }^{\prime} C^{n}$ is the mean position and $A\, \&\, B$ are extreme positions
$\Rightarrow x_{\max }=\frac{2 a}{b}$
$\&\, V_{\max }=\omega A=\sqrt{b} \cdot \frac{a}{b}=\frac{a}{\sqrt{b}}$