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Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 3.2 : Interpretation of the Derivative
9. The position of an object at any time \(t\) is given by \(\displaystyle s\left( t \right) = \frac{{t + 1}}{{t + 4}}\).
- Determine the velocity of the object at any time \(t\).
- Does the object ever stop moving? If yes, at what time(s) does the object stop moving?
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a Determine the velocity of the object at any time \(t\). Show SolutionWe know that the derivative of a function gives is the velocity of the object and so we’ll first need the derivative of this function. We computed this derivative in Problem 9 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding.
Note that we did use a different letter for the function in the previous problem, but the work is identical. So, from our previous work we know that the derivative is,
\[s'\left( t \right) = \frac{3}{{{{\left( {t + 4} \right)}^2}}}\]b Does the object ever stop moving? If yes, at what time(s) does the object stop moving? Show Solution
We know that the object will stop moving if the velocity (i.e. the derivative) is zero. In this case the derivative is a rational expression and clearly the numerator will never be zero. Therefore, the derivative will not be zero and therefore the object never stops moving.