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:

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

Answer 1

Answer: (A)

Given that acceleration of car is
$f=a-bx$
For maximum velocity, acceleration should be zero.
$\Rightarrow a-bx=0$
$\Rightarrow x=\frac{a}{b}$
At $x=\frac{a}{b}$, the particle has its maximum velocity.
We use $f=\frac{vdv}{dx}=a-bx$
$\Rightarrow \frac{v^2}{2}=ax-\frac{b{x}^2}{2}+c$
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 $B$
$\Rightarrow ax-\frac{b{x}^2}{2}=0$
$\Rightarrow x=0;x=\left(\frac{2a}{b}\right)$
$\Rightarrow$ Distance between the stations is $\frac{2a}{b}$.
Alternate $:f=a-bx$ means particle will execute 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
$\Rightarrow x_{\max }=\frac{2a}{b}$
$\&V_{\max }=\omega A=\sqrt{b}\cdot \frac{a}{b}=\frac{a}{\sqrt{b}}$.