Question

The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in seconds. What is the car’s maximum acceleration on the time interval 0≤t≤6 ?

Answer 1

Answer:

The maximum acceleration over that interval is $A(6) = 28$.

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable $t$.

Notice that the interval of interest $0 \le t \le 6$ is closed on both ends. In other words, this interval includes both endpoints: $t = 0$ and $t= 6$. Over this interval, the value of $A(t)$ might be maximized when $t$ is at the following:

• One of the two endpoints of this interval, where $t = 0$ or $t = 6$.
• A local maximum of $A(t)$, where $A^\prime(t) = 0$ (first derivative of $A(t)\!$ is zero) and $A^{\prime\prime}(t)$ (second derivative of $\! A(t)$ is smaller than zero.)

Start by calculating the value of $A(t)$ at the two endpoints:

• $A(0) = 10$.
• $A(6) = 28$.

Apply the power rule to find the first and second derivatives of $A(t)$:

$\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}$.

$\displaystyle A^{\prime\prime}(t) = 6\, t - 15$.

Notice that both $t = 1$ and $t = 4$ are first derivatives of $A^{\prime}(t)$ over the interval $0 \le t \le 6$.

However, among these two zeros, only $t = 1\!$ ensures that the second derivative $A^{\prime\prime}(t)$ is smaller than zero (that is: $A^{\prime\prime}(1) < 0$.) If the second derivative $A^{\prime\prime}(t)\!$ is non-negative, that zero of $A^{\prime}(t)$ would either be an inflection point (if$A^{\prime\prime}(t) = 0$) or a local minimum (if $A^{\prime\prime}(t) > 0$.)

Therefore $\! t = 1$ would be the only local maximum over the interval $0 \le t \le 6\!$.

Calculate the value of $A(t)$ at this local maximum:

• $A(1) = 15.5$.

Compare these three possible maximum values of $A(t)$ over the interval $0 \le t \le 6$. Apparently, $t = 6$ would maximize the value of $A(t)\!$. That is: $A(6) = 28$ gives the maximum value of $\! A(t)$ over the interval $0 \le t \le 6\!$.

However, note that the maximum over this interval exists because $t = 6\!$ is indeed part of the $0 \le t \le 6$ interval. For example, the same $A(t)$ would have no maximum over the interval $0 \le t < 6$ (which does not include $t = 6$.)