Question

The displacement of a body at any time t is given by \[ x = -\frac{3}{4}t^2 + 12t + 3 \] The velocity of the body is zero after:

Hint:

Always double-check the signs when differentiating quadratic displacement equations. A negative $t^2$ term indicates constant deceleration, meaning the object will inevitably stop and reverse its direction.

Answer 1

Step 1: Understanding the Concept:
In classical kinematics, the instantaneous velocity of a moving body is defined precisely as the time derivative of its position or displacement vector.
To find the exact time when the body briefly comes to rest, we must calculate the velocity function and explicitly set it equal to zero.
Step 2: Key Formula or Approach:
The instantaneous velocity $v$ is derived from the displacement $x$ using basic differential calculus: $v = \frac{dx}{dt}$.
Once the generalized expression for $v$ is obtained, solve the algebraic equation $v(t) = 0$ for time $t$.
Step 3: Detailed Explanation:
The given displacement function with respect to time is:
\[ x = -\frac{3}{4}t^2 + 12t + 3 \] Differentiate this entire equation with respect to $t$ to find the instantaneous velocity $v$:
\[ v = \frac{d}{dt} \left( -\frac{3}{4}t^2 + 12t + 3 \right) \] Applying basic power rule differentiation:
\[ v = -\frac{3}{4}(2t) + 12(1) + 0 \] \[ v = -\frac{3}{2}t + 12 \] To precisely find when the velocity becomes absolute zero, we set $v = 0$:
\[ -\frac{3}{2}t + 12 = 0 \] Rearranging the simple linear equation:
\[ \frac{3}{2}t = 12 \] \[ t = 12 \times \frac{2}{3} \] \[ t = \frac{24}{3} = 8\text{ s} \] Step 4: Final Answer:
The velocity of the moving body drops to zero exactly after $8\text{ s}$.