The displacement of a particle moving in a straight line is given by the expression $x=A t^3+B t^2+C t+D$ in…
The displacement of a particle moving in a straight line is given by the expression $x=A t^3+B t^2+C t+D$ in metres, where $t$ is in seconds and $A, B, C$ and $D$ are constants. The ratio between the initial acceleration and initial velocity is
$\frac{2 C}{B}$
$\frac{2 B}{C}$
$2 \mathrm{C}$
$\frac{C}{2 B}$
Solution
Displacement of a particle moving in a straight line $(x)=A t^3+B t^2+C t+D$
$\therefore$ Velocity,$\quad v=\frac{d x}{d t}=3 A t^2+2 B t+C$
$
v_{\text {initial }}=C
$
and
$
\begin{aligned}
a & =\frac{d^2 x}{d t^2}=6 A t+2 B \\
a_{\text {initial }} & =2 B
\end{aligned}
$
$\therefore$ Ratio between initial acceleration and initial velocity is $\frac{2 B}{C}$
Asked in: JEE Mains - Motion In One Dimension - Test 3