Solved on Sep 12, 2023

Find the interval(s) where $s(t) = -\frac{t^{3}}{3} + \frac{7t^{2}}{2} - 6t + 2$ describes a slowing hummingbird, where $t \geq 0$ is time in seconds and $s$ is position in yards.

STEP 1

Assumptions1. The position of the hummingbird is given by the function $s(t)=-\frac{t^{3}}{3}+\frac{7 t^{}}{}-6 t+$ . $t \geq0$ is measured in seconds and $s$ is measured in yards3. We need to find the interval(s) where the hummingbird is slowing down

STEP 2

To find where the hummingbird is slowing down, we need to find where the velocity and acceleration have opposite signs. The velocity is the derivative of the position function, and the acceleration is the derivative of the velocity function.
First, let's find the velocity function by taking the derivative of the position function.
$v(t) = s'(t)$

STEP 3

Now, plug in the given position function to calculate the derivative.
$v(t) = s'(t) = -t^{2}+7t-6$

STEP 4

Next, let's find the acceleration function by taking the derivative of the velocity function.
$a(t) = v'(t)$

STEP 5

Now, plug in the given velocity function to calculate the derivative.
$a(t) = v'(t) = -2t+7$

STEP 6

We need to find the intervals where the velocity and acceleration have opposite signs. This occurs when the velocity is positive and the acceleration is negative, or when the velocity is negative and the acceleration is positive.
First, let's find the critical points of the velocity and acceleration functions by setting them equal to zero and solving for $t$ .
$-t^{2}+t-6=0$ $-2t+=0$

STEP 7

olve the equations for $t$ .
$-t^{2}+7t-6=0 \Rightarrow t =1,6$ $-2t+7=0 \Rightarrow t = \frac{7}{2}$

STEP 8

Now, let's test the intervals determined by the critical points in the velocity and acceleration functions. The intervals are $[0,1]$ , $(1, \frac{7}{2})$ , $(\frac{7}{2},6)$ , and $(6, \infty)$ .
We choose test points in these intervals and plug them into the velocity and acceleration functions. If the signs are opposite, then the hummingbird is slowing down in that interval.

STEP 9

Test the intervals in the velocity and acceleration functions.
For $[,]$ Choose $t=.5$ , then $v(.5) >$ and $a(.5) <$ . So, the hummingbird is slowing down on this interval.
For $(, \frac{7}{2})$ Choose $t=2$ , then $v(2) >$ and $a(2) >$ . So, the hummingbird is not slowing down on this interval.
For $(\frac{7}{2},6)$ Choose $t=4$ , then $v(4) <$ and $a(4) >$ . So, the hummingbird is slowing down on this interval.
For $(6, \infty)$ Choose $t=7$ , then $v(7) <$ and $a(7) <$ . So, the hummingbird is not slowing down on this interval.

STEP 10

From the above analysis, the hummingbird is slowing down on the intervals $[0,]$ and $(\frac{7}{2},6)$ .
So, the answer is $[0,] \cup (\frac{7}{2},6)$ .

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Find the interval(s) where $s(t) = -\frac{t^{3}}{3} + \frac{7t^{2}}{2} - 6t + 2$ describes a slowing hummingbird, where $t \geq 0$ is time in seconds and $s$ is position in yards.

STEP 1

Assumptions1. The position of the hummingbird is given by the function $s(t)=-\frac{t^{3}}{3}+\frac{7 t^{}}{}-6 t+$ . $t \geq0$ is measured in seconds and $s$ is measured in yards3. We need to find the interval(s) where the hummingbird is slowing down

STEP 2

STEP 3

Now, plug in the given position function to calculate the derivative.
$v(t) = s'(t) = -t^{2}+7t-6$

STEP 4

Next, let's find the acceleration function by taking the derivative of the velocity function.
$a(t) = v'(t)$

STEP 5

Now, plug in the given velocity function to calculate the derivative.
$a(t) = v'(t) = -2t+7$

STEP 6

STEP 7

olve the equations for $t$ .
$-t^{2}+7t-6=0 \Rightarrow t =1,6$ $-2t+7=0 \Rightarrow t = \frac{7}{2}$

STEP 8

STEP 9

STEP 10

From the above analysis, the hummingbird is slowing down on the intervals $[0,]$ and $(\frac{7}{2},6)$ .
So, the answer is $[0,] \cup (\frac{7}{2},6)$ .

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Find the interval(s) where s(t)=−t33+7t22−6t+2s(t) = -\frac{t^{3}}{3} + \frac{7t^{2}}{2} - 6t + 2s(t)=−3t3​+27t2​−6t+2 describes a slowing hummingbird, where t≥0t \geq 0t≥0 is time in seconds and sss is position in yards.

STEP 1

STEP 2

STEP 3

Now, plug in the given position function to calculate the derivative.v(t)=s′(t)=−t2+7t−6v(t) = s'(t) = -t^{2}+7t-6v(t)=s′(t)=−t2+7t−6

STEP 4

Next, let's find the acceleration function by taking the derivative of the velocity function.a(t)=v′(t)a(t) = v'(t)a(t)=v′(t)

STEP 5

Now, plug in the given velocity function to calculate the derivative.a(t)=v′(t)=−2t+7a(t) = v'(t) = -2t+7a(t)=v′(t)=−2t+7

STEP 6

STEP 7

olve the equations for ttt.−t2+7t−6=0⇒t=1,6-t^{2}+7t-6=0 \Rightarrow t =1,6−t2+7t−6=0⇒t=1,6−2t+7=0⇒t=72-2t+7=0 \Rightarrow t = \frac{7}{2}−2t+7=0⇒t=27​

STEP 8

STEP 9

STEP 10

From the above analysis, the hummingbird is slowing down on the intervals [0,][0,][0,] and (72,6)(\frac{7}{2},6)(27​,6).So, the answer is [0,]∪(72,6)[0,] \cup (\frac{7}{2},6)[0,]∪(27​,6).

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Find the interval(s) where $s(t) = -\frac{t^{3}}{3} + \frac{7t^{2}}{2} - 6t + 2$ describes a slowing hummingbird, where $t \geq 0$ is time in seconds and $s$ is position in yards.

Now, plug in the given position function to calculate the derivative.
$v(t) = s'(t) = -t^{2}+7t-6$

Next, let's find the acceleration function by taking the derivative of the velocity function.
$a(t) = v'(t)$

Now, plug in the given velocity function to calculate the derivative.
$a(t) = v'(t) = -2t+7$

olve the equations for $t$ .
$-t^{2}+7t-6=0 \Rightarrow t =1,6$ $-2t+7=0 \Rightarrow t = \frac{7}{2}$

From the above analysis, the hummingbird is slowing down on the intervals $[0,]$ and $(\frac{7}{2},6)$ .
So, the answer is $[0,] \cup (\frac{7}{2},6)$ .