A body starts from the origin and moves along | vedclass

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(B) The position of the body is given by the equation $x = 4t^2 - 12t$.To find the velocity $v$,we differentiate the position $x$ with respect to time

Vedclass · Accepted Answer

$8$

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(B) The position of the body is given by the equation $x = 4t^2 - 12t$.To find the velocity $v$,we differentiate the position $x$ with respect to time $t$:$v = \frac{dx}{dt} = \frac{d}{dt}(4t^2 - 12t) = 8t - 12 \, m/s$.To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:$a = \frac{dv}{dt} = \frac{d}{dt}(8t - 12) = 8 \, m/s^2$.Therefore,the acceleration of the particle is $8 \, m/s^2$.

$A$ body starts from the origin and moves along the $x$-axis such that its position at any instant is $x = 4t^2 - 12t$,where $t$ is in seconds and $x$ is in meters. What is the acceleration of the particle in $m/s^2$?

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