The acceleration at the end of 2 s of a part | vedclass

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(A) Given the displacement equation: $S = 4t^3 - 8t^2 + 5t + 4$. Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(A) Given the displacement equation: $S = 4t^3 - 8t^2 + 5t + 4$. Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dS}{dt} = \frac{d}{dt}(4t^3 - 8t^2 + 5t + 4) = 12t^2 - 16t + 5$. Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(12t^2 - 16t + 5) = 24t - 16$. To find the acceleration at $t = 2 \ s$,substitute $t = 2$ into the acceleration equation: $a = 24(2) - 16 = 48 - 16 = 32 \ m \ s^{-2}$.

$t (s)$	$0$	$1$	$2$	$3$
$x (m)$	$-2$	$0$	$6$	$16$

The acceleration at the end of $2 \ s$ of a particle whose motion is represented by the equation $S = 4t^3 - 8t^2 + 5t + 4$ is $......$ (in $m \ s^{-2}$)

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