Question

Particle is moving in a straight path with an acceleration $a=b{t}^n$, where b is a positive constant $(n\ne -1)$. The relation between the velocity and the position vector of the particle at time t = 1s [Assuming particle is in rest at origin at t=0]

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

Answer 1

The correct option is B $V=(n+2)r$

Given that,

$a=b{t}^n$

$\frac{dv}{dt}=b{t}^n$

$dv=b{t}^{n}dt$

$v=b\frac{t^{n+1}}{n+1}........\left(1\right)$

$\frac{dr}{dt}=b\frac{t^{n+1}}{n+1}$

$dr=\frac{b}{n+1}\int t^{n+1}dt$

$r=\frac{b}{n+1}\frac{t^{n+2}}{n+2}$

$r=\frac{Vt}{n+2}\left(from1\right)$

$att=1s$

$V=(n+2)r$