To find the velocity of the object in terms of time, we need to take the derivative of the displacement function \(s(t)\) with respect to time \(t\). The displacement function is given by: \[ s(t) = 8t^3 - 3t^2 - 4 \] The velocity \(v(t)\) is the first derivative of the displacement \(s(t)\) with respect to time \(t\): \[ v(t) = \frac{ds(t)}{dt} \] Let's differentiate \(s(t)\): \[ s(t) = 8t^3 - 3t^2 - 4 \] Using the power rule for differentiation, which states that \(\frac{d}{dt}[t^n] = nt^{n-1}\), we get: \[ \frac{d}{dt}[8t^3] = 8 \cdot 3t^{3-1} = 24t^2 \] \[ \frac{d}{dt}[-3t^2] = -3 \cdot 2t^{2-1} = -6t \] \[ \frac{d}{dt}[-4] = 0 \] Combining these results, we get: \[ v(t) = 24t^2 - 6t \] Therefore, the correct answer is: \[ \boxed{B} \]

To find the velocity of an object in terms of time, we need to take the derivative of the displacement function $s(t)$ with respect to time $t$. The displacement function is given by:

$s(t) = 8t^3 - 3t^2 - 4$

The velocity $v(t)$ is the first derivative of the displacement $s(t)$ with respect to time $t$:

$v(t) = \frac{ds(t)}{dt}$

Now, let's differentiate $s(t)$:

$s(t) = 8t^3 - 3t^2 - 4$

Using the power rule for differentiation, which states that $\frac{d}{dt}[t^n] = nt^{n-1}$, we get:

$\frac{d}{dt}[8t^3] = 8 \cdot 3t^{3-1} = 24t^2$

$\frac{d}{dt}[-3t^2] = -3 \cdot 2t^{2-1} = -6t$

$\frac{d}{dt}[-4] = 0$

Combining these results, we get:

$v(t) = 24t^2 - 6t$

Therefore, the velocity of the object in terms of time is:

$v(t) = 24t^2 - 6t$