Question

The displacement of a particle moving in a straight line depends on time as
$x=\alpha t^{3}+\beta t^{2}+\gamma t+\delta$
The ratio of initial acceleration to its initial velocity depends :

• A. only on $\alpha$ and $\beta$
• B. only on $\beta$ and $\gamma$
• C. only on $\alpha$ and $\beta$
• D. only on $\alpha$

Answer 1

Answer: (B)

Rate of change of displacement is velocity and rate of change of velocity is acceleration.
The given equation for displacement is
$x=\alpha t^{3}+\beta t^{2}+\gamma t+\delta$
We know that velocity $=\frac{d x}{d t}$
$\therefore v=\frac{d x}{d t}=3 t^{2} \alpha+2 t \beta+\gamma$
Initial velocity when $t=0$, is
$v_{0}=3 \alpha \times 0+2 \beta \times 0+\gamma=\gamma$ ...(i)
Also acceleration $=\frac{d v}{d t}$
$\therefore a=\frac{d v}{d t}=\frac{d^{2} x}{d t^{2}}=6 t \alpha+2 \beta$
Initial acceleration when $t=0$, is
$a_{0}=6 t \times 0+2 \beta=2 \beta$ ...(ii)
Ratio of initial acceleration to initial velocity is
$\frac{a_{0}}{v_{0}}=\frac{2 \beta}{\gamma}$
which shown that this ratio depends only on $\beta$ and $\gamma$.