Answer:
The maximum acceleration over that interval is .
Step-by-step explanation:
The acceleration of this car is modelled as a function of the variable .
Notice that the interval of interest is closed on both ends. In other words, this interval includes both endpoints: and . Over this interval, the value of might be maximized when is at the following:
- One of the two endpoints of this interval, where or .
- A local maximum of , where (first derivative of is zero) and (second derivative of is smaller than zero.)
Start by calculating the value of at the two endpoints:
- .
- .
Apply the power rule to find the first and second derivatives of :
.
.
Notice that both and are first derivatives of over the interval .
However, among these two zeros, only ensures that the second derivative is smaller than zero (that is: .) If the second derivative is non-negative, that zero of would either be an inflection point (if) or a local minimum (if .)
Therefore would be the only local maximum over the interval .
Calculate the value of at this local maximum:
- .
Compare these three possible maximum values of over the interval . Apparently, would maximize the value of . That is: gives the maximum value of over the interval .
However, note that the maximum over this interval exists because is indeed part of the interval. For example, the same would have no maximum over the interval (which does not include .)