Question
\[
d(t)=\frac{t^2-6}{2t^3}, \qquad t>0
\]
Find the time(s) when the object is stationary.
You must use calculus and show any derivatives that you need when solving this problem.
Try the question yourself first. If you get stuck, open the hints before using the full walkthrough.
Step 1
Choose the rule
The displacement function is a fraction, so which differentiation rule is the best starting point?
Step 2
Choose \(u\) and \(v\)
Write \(d(t)\) as \(\frac{u}{v}\). Which choice matches the function?
Step 3
Differentiate the top and bottom
Find \(u'\) and \(v'\).
Step 4
Apply the quotient rule
Use \(\frac{d}{dt}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}\).
Step 5
Simplify and solve for stationary times
An object is stationary when its velocity is zero, so set \(d'(t)=0\).
\[
d'(t)=\frac{4t^4-6t^2(t^2-6)}{4t^6}
\]
\[
d'(t)=\frac{-2t^4+36t^2}{4t^6}
\]
\[
d'(t)=\frac{-2t^2(t^2-18)}{4t^6}
\]
Which time value is valid when \(t>0\)?
Step 6
Final answer and graph
The object is stationary at:
\[
t=\sqrt{18}=3\sqrt{2}\text{ seconds}
\]
Interactive sketch
This graph shows the displacement function for \(t>0\), with the stationary time marked so you can see where the tangent becomes horizontal.