Question

A particle is moving in a straight path with an acceleration $a=bn$, where $b$ is a positive constant $(n\ne -1)$. Assuming particle is at rest at origin at $t=0$, the relation between the velocity and the position$\left(r\right)$ of the particle at time $t=1\text{s}$ is

(Assume SI units)

• A. $v=r$
• B. $v=2r$
• C. $v=\frac{(n+3)r}{2}$
• D. $v=\frac{(n+1)r}{2}$

Answer 1

The correct option is B $v=2r$

Given acceleration $a=\frac{dv}{dt}=bn$
$\underset{0}{\overset{v}{\int }}dv=bn\underset{0}{\overset{t}{\int }}dt$ (at $t=0,v=0$)
$v=bnt$ ... (1)
again,
$v=\frac{dx}{dt}=bnt$
$\underset{0}{\overset{x}{\int }}dx=bn\underset{0}{\overset{t}{\int }}tdt$ (at $t=0,x=0$)
$x=bn\times \frac{t^2}{2}$ ... (2)
From (1) and (2),
$\frac{v}{x}=\frac{bnt}{bn{t}^2/2}=\frac{2}{t}$
At $t=1,\text{let}x=r$ ,
then $\frac{v}{r}=\frac{2}{1}=2$
i.e., $v=2r$