Question

A bus moves from stop $A$ to the next stop $B$. Its acceleration varies as $a=\alpha -\beta x$, where $\alpha$ and $\beta$ are positive constants and $x$ is the distance from the stop $A$ to stop $B$. The distance between $A$ and $B$ and the maximum speed of the bus are

• A. $\frac{2\alpha }{\beta }\text{and}\frac{\alpha }{\sqrt{\beta }}$
• B. $\frac{\alpha }{\beta }\text{and}\frac{\alpha }{\sqrt{\beta }}$
• C. $\frac{\alpha }{\beta }\text{and}\frac{2\alpha }{\sqrt{\beta }}$
• D. $\frac{2\alpha }{\beta }\text{and}\frac{2\alpha }{\sqrt{\beta }}$

Answer 1

The correct option is A $\frac{2\alpha }{\beta }\text{and}\frac{\alpha }{\sqrt{\beta }}$

The acceleration equation $a=\alpha -\beta x$ can be written as $vdv=(\alpha -\beta x)dx$ The bus starts from rest at the stop $A$. By integrating, we get
${\int }_0^{v}vdv={\int }_0^x(\alpha -\beta x)dx$
$v^2=2\alpha x-\beta x^2$
Substitute $v=0$ (the bus comes to rest at the stop B) to get the distance between two stops.
$\therefore$ Distance between two stops = $x=\frac{2\alpha }{\beta }$


The velocity is maximum when acceleration (first derivative of velocity) is zero. Substitute $a=0$ in $a=\alpha -\beta x$ to get $x=\frac{\alpha }{\beta }$, the position of maximum velocity.
Substitute $x=\frac{\alpha }{\beta }$ in $v^2=2\alpha x-\beta x^2$ to get the maximum velocity $v=\frac{\alpha }{\sqrt{\beta }}$