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(C) The distance of the particle is given by $S = f(t) = t^3 - 5t^2 + 8t$.Velocity $v(t)$ is the first derivative of distance with respect to time $t$

Vedclass · Accepted Answer

$20$

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(C) The distance of the particle is given by $S = f(t) = t^3 - 5t^2 + 8t$.Velocity $v(t)$ is the first derivative of distance with respect to time $t$:$v(t) = \frac{dS}{dt} = \frac{d}{dt}(t^3 - 5t^2 + 8t) = 3t^2 - 10t + 8$.Acceleration $a(t)$ is the derivative of velocity with respect to time $t$:$a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 10t + 8) = 6t - 10$.To find the acceleration at $t = 5 	ext{ sec}$, substitute $t = 5$ into the acceleration equation:$a(5) = 6(5) - 10 = 30 - 10 = 20 	ext{ cm/sec}^2$.Thus, the acceleration of the particle at $t = 5 	ext{ sec}$ is $20 	ext{ cm/sec}^2$.

If a particle is moving in a straight line so that after $t$ seconds its distance $S$ (in $cm$) from a fixed point on the line is given by $S = f(t) = t^3 - 5t^2 + 8t$, then the acceleration of the particle at $t = 5 \text{ sec}$ is (in $cm/sec^2$).

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