Mathematics

Mathematics

Problem

For a particle moving in a straight line, the velocity for 0≤t≤5 s0 \le t \le 5\,\text{s} is given by v(t)={2t2,0≤t≤3 s18−6t,3<t≤5 sv(t)=\begin{cases}2t^{2}, & 0 \le t \le 3\,\text{s}\\18 - 6t, & 3 < t \le 5\,\text{s}\end{cases} What is the instantaneous acceleration just after t=3 st = 3\,\text{s}? a) 12 m s−212\,\text{m s}^{-2} b) 6 m s−26\,\text{m s}^{-2} c) −6 m s−2-6\,\text{m s}^{-2} d) −12 m s−2-12\,\text{m s}^{-2}
Intuitive explanation

The key here is recognizing that "just after t=3 st = 3\,\text{s}" means you need the right-hand derivative—the instantaneous rate of change of velocity using the velocity formula that applies immediately past t=3t = 3. Acceleration is simply dvdt\frac{dv}{dt}, so find the derivative of the appropriate piece and evaluate it at t=3t = 3.

Intuitive explanation

The key here is recognizing that "just after t=3 st = 3\,\text{s}" means you need the right-hand derivative—the instantaneous rate of change of velocity using the velocity formula that applies immediately past t=3t = 3. Acceleration is simply dvdt\frac{dv}{dt}, so find the derivative of the appropriate piece and evaluate it at t=3t = 3.

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