Mathematics, 10.01.2021 14:00 emmanuellugo40

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 ?

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Answer from: avahymowitz44

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 .

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 is at the following:

One of the two endpoints of this interval, where t = 0

.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:

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:

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 .)

Answer from: Quest

step-by-step explanation:

how many centimeters did the level of the salsa go down by?

Answer from: Quest

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,

0 = ( 2 - 4 ) ( p + q )0 = ( 2 - 4 ) ( 2 + 5 i ), by multiplying these0 = ( 4 + 10 i - 8 - 20 i )0 = 4 - 8 + 10 i - 20 i0 = - 4 -10 i 0 = 2 ( - 2 - 5 i )0 / 2 = - 2 - 5 i- 2 - 5 i = 0- 5 i = 2i = 2 / - 5

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The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+...