Question

For a particle moving along a straight line, the displacement 'x' depends on time 't' as $x=\alpha t^3+\beta t^2+\gamma t+\delta$. The ratio of its initial acceleration to its initial velocity depends

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

Answer 1

Answer: (B)

only on $\beta$ and $\gamma$

Given,

Displacement, $x=\alpha t^3+\beta t^2+\gamma t+\delta$

Velocity, $v=\frac{dx}{dt}=3\alpha t^2+2\beta t+\gamma$

The initial velocity, $t=0sec$

$v_0=\frac{dx}{dt}=\gamma$. . . . . .(1)

Acceleration, $a=\frac{d^{2}x}{d{t}^2}=\frac{dv}{dt}=6\alpha t+2\beta$

The Initial acceleration at $t=0sec$

$a_0=\frac{dv}{dt}=2\beta$. . . . . . . . .(2)

Ratio, $\frac{a_0}{v_0}=\frac{2\beta }{\gamma }$

The ratio of initial acceleration to initial velocity depends on $\beta$ and $\gamma$ only.

The correct option is B.