Question

The displacement (s) of a particle moving along a straight line is related to time $t$ as $s=a{t}^3+b{t}^3+ct$, where $a,b$ and $c$ are constants. What is the ratio of its initial velocity and initial acceleration?

• A. $\infty$
• B. $\frac{b}{2c}$
• C. $\frac{c}{2c}$
• D. $\frac{c}{b}$

Answer 1

The correct option is A $\infty$

Given,

$s=a{t}^3+b{t}^3+ct$

Initial velocity at $t=0s$ is

$v=\frac{ds}{dt}$

$v=3a{t}^2+3b{t}^2+c$

$v{|}_{t=0s}=3a(0{)}^2+3b(0{)}^2+c$

$v{|}_{t=0s}=c$

Initial acceleration at $t=0s$ is

$a=\frac{dv}{dt}$

$a=6at+6bt$

$a{|}_{t=0s}=6a\left(0\right)+6b\left(0\right)$

$a{|}_{t=0s}=0$

Ratio, $\frac{v{|}_{t=0s}}{a{|}_{t=0s}}=\frac{c}{0}=\infty$