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(B) The displacement of the particle is given by $x = A t^3 + B t^2 + C t + D$. To find the velocity $v$, we differentiate $x$ with respect to time $t

Vedclass · Accepted Answer

$\frac{2 B}{C}$

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(B) The displacement of the particle is given by $x = A t^3 + B t^2 + C t + D$. To find the velocity $v$, we differentiate $x$ with respect to time $t$: $v = \frac{dx}{dt} = 3At^2 + 2Bt + C$. The initial velocity $(v_{	ext{initial}})$ is the velocity at $t = 0$: $v_{	ext{initial}} = 3A(0)^2 + 2B(0) + C = C$. To find the acceleration $a$, we differentiate $v$ with respect to time $t$: $a = \frac{dv}{dt} = 6At + 2B$. The initial acceleration $(a_{	ext{initial}})$ is the acceleration at $t = 0$: $a_{	ext{initial}} = 6A(0) + 2B = 2B$. The ratio between the initial acceleration and initial velocity is $\frac{a_{	ext{initial}}}{v_{	ext{initial}}} = \frac{2B}{C}$.

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

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