The displacement of a particle moving in a st | vedclass

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(B) The displacement is given by $x = \alpha t^3 + \beta t^2 + \gamma t + \delta$. Velocity $v$ is the derivative of displacement with respect to time

Vedclass · Accepted Answer

only on $\beta$ and $\gamma$

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(B) The displacement is given by $x = \alpha t^3 + \beta t^2 + \gamma t + \delta$. Velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = 3\alpha t^2 + 2\beta t + \gamma$. Initial velocity $(v_i)$ at $t = 0$ is $v_i = \gamma$. Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = 6\alpha t + 2\beta$. Initial acceleration $(a_i)$ at $t = 0$ is $a_i = 2\beta$. The ratio of initial acceleration to initial velocity is $\frac{a_i}{v_i} = \frac{2\beta}{\gamma}$. Thus,the ratio depends only on $\beta$ and $\gamma$.

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 on:

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