Question

For a particle moving along a straight line, the position $x$ depends on time $t$ as $x=A{t}^3+B{t}^2+Ct+D$. The ratio of its initial velocity to its initial acceleration will be

• A. $\frac{C}{B}$
• B. $\frac{D}{C}$
• C. $\frac{C}{2B}$
• D. $\frac{2B}{3A}$

Answer 1

The correct option is C $\frac{C}{2B}$

As given: $x=A{t}^3+B{t}^2+Ct+D$
For velocity, $v=\frac{dx}{dt}=3A{t}^2+2Bt+C$
For initial velocity $t=0$ and $v=v_0$ put in above equation
$v_0=0+0+C$
$\Rightarrow v_0=C$
For acceleration, $a=\frac{dv}{dt}=6At+2B$
For initial acceleration $t=0$ and $a=a_0$ put in above equation
$a_0=0+2B$
$\Rightarrow a_0=2B$
The ratio of initial velocity to initial acceleration $=\frac{v_0}{a_0}=\frac{C}{2B}$