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

\frac{C}{B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{D}{C}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{C}{2 B}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{2 B}{3 A}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{C}{2 B}$
As given: $x = A t^{3} + B t^{2} + C t + D$
For velocity, $v = \frac{d x}{d t} = 3 A t^{2} + 2 B t + C$
For initial velocity $t = 0$ and $v = v_{0}$ put in above equation
$v_{0} = 0 + 0 + C$
$\Rightarrow v_{0} = C$
For acceleration, $a = \frac{d v}{d t} = 6 A t + 2 B$
For initial acceleration $t = 0$ and $a = a_{0}$ put in above equation
$a_{0} = 0 + 2 B$
$\Rightarrow a_{0} = 2 B$
The ratio of initial velocity to initial acceleration $= \frac{v_{0}}{a_{0}} = \frac{C}{2 B}$

Suggest Corrections

Similar questions

x

t

x = A t^{3} + B t^{2} + C t + D

x = α t^{3} + β t^{2} + γ t + δ

x -

x = {a t}^{3} + {b t}^{2} + c t + d

t

s = a t^{3} + b t^{3} + c t

a, b

c

x = - 5 m

x = - 11 m

2 s

Join BYJU'S Learning Program