(B) The displacement is given by $s = t^{3} - 4t^{2} - 5t$. Velocity $v$ is the rate of change of displacement with respect to time,given by $v = \fra

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(B) The displacement is given by $s = t^{3} - 4t^{2} - 5t$. Velocity $v$ is the rate of change of displacement with respect to time,given by $v = \frac{ds}{dt}$. Differentiating $s$ with respect to $t$: $\frac{ds}{dt} = \frac{d}{dt}(t^{3} - 4t^{2} - 5t) = 3t^{2} - 8t - 5$. To find the velocity at $t = 2 \text{ sec}$,substitute $t = 2$ into the expression for $v$: $v = 3(2)^{2} - 8(2) - 5$. $v = 3(4) - 16 - 5$. $v = 12 - 16 - 5$. $v = -9 \text{ units/sec}$.

Class 12 Mathematics Application of Derivatives Rate of Change of Quantities

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The displacement of a particle at time $t$ is $s = t^{3} - 4t^{2} - 5t$. The velocity of the particle at $t = 2 \text{ sec}$ is:

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A
$\left(\frac{1}{9}\right) \text{ units/sec}$
B
$-9 \text{ units/sec}$
C
$9 \text{ units/sec}$
D
$\left(-\frac{1}{9}\right) \text{ units/sec}$

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The displacement of a particle at time $t$ is $s = t^{3} - 4t^{2} - 5t$. The velocity of the particle at $t = 2 \text{ sec}$ is:

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