QuestionFind the interval(s) where the car slows down given $s(t)=-\frac{2 t^{3}}{3}+3 t^{2}+20 t-3$ , $t \geq 0$ .

Studdy Solution

STEP 1

Assumptions1. The position of the car is given by the function $s(t)=-\frac{ t^{3}}{3}+3 t^{}+20 t-3$ . $t \geq0$ is measured in seconds and $s$ is measured in meters3. We need to find the interval(s) where the car is slowing down

STEP 2

A car is slowing down when its velocity and acceleration have opposite signs. First, we need to find the velocity, which is the derivative of the position function.
$v(t) = s'(t)$

STEP 3

Now, calculate the derivative of the position function.
$v(t) = -2t^{2} +6t +20$

STEP 4

Next, we need to find the acceleration, which is the derivative of the velocity function.
$a(t) = v'(t)$

STEP 5

Now, calculate the derivative of the velocity function.
$a(t) = -4t +$

STEP 6

To find where the car is slowing down, we need to find where the velocity and acceleration have opposite signs. This occurs where the velocity is positive and the acceleration is negative, or where the velocity is negative and the acceleration is positive.First, find the critical points of the velocity and acceleration functions by setting them equal to zero and solving for $t$ .
$v(t) =0, a(t) =0$

STEP 7

olve the equation $v(t) =0$ for $t$ .
$-2t^{2} +6t +20 =0$

STEP 8

olve the equation $a(t) =0$ for $t$ .
$-4t +6 =0$

STEP 9

The solutions to these equations will give us the critical points, which divide the time axis into intervals. We can then test these intervals to see where the velocity and acceleration have opposite signs.

STEP 10

Now, we need to test the intervals between the critical points to see where the car is slowing down. We do this by choosing a test point in each interval and evaluating the sign of the velocity and acceleration at these points.

STEP 11

If the signs of the velocity and acceleration are opposite at a test point, then the car is slowing down on that interval. Write these intervals in interval notation.

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QuestionFind the interval(s) where the car slows down given $s(t)=-\frac{2 t^{3}}{3}+3 t^{2}+20 t-3$ , $t \geq 0$ .

Studdy Solution

STEP 1

Assumptions1. The position of the car is given by the function $s(t)=-\frac{ t^{3}}{3}+3 t^{}+20 t-3$ . $t \geq0$ is measured in seconds and $s$ is measured in meters3. We need to find the interval(s) where the car is slowing down

STEP 2

A car is slowing down when its velocity and acceleration have opposite signs. First, we need to find the velocity, which is the derivative of the position function.
$v(t) = s'(t)$

STEP 3

Now, calculate the derivative of the position function.
$v(t) = -2t^{2} +6t +20$

STEP 4

Next, we need to find the acceleration, which is the derivative of the velocity function.
$a(t) = v'(t)$

STEP 5

Now, calculate the derivative of the velocity function.
$a(t) = -4t +$

STEP 6

STEP 7

olve the equation $v(t) =0$ for $t$ .
$-2t^{2} +6t +20 =0$

STEP 8

olve the equation $a(t) =0$ for $t$ .
$-4t +6 =0$

STEP 9

The solutions to these equations will give us the critical points, which divide the time axis into intervals. We can then test these intervals to see where the velocity and acceleration have opposite signs.

STEP 10

Now, we need to test the intervals between the critical points to see where the car is slowing down. We do this by choosing a test point in each interval and evaluating the sign of the velocity and acceleration at these points.

STEP 11

If the signs of the velocity and acceleration are opposite at a test point, then the car is slowing down on that interval. Write these intervals in interval notation.

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QuestionFind the interval(s) where the car slows down given s(t)=−2t33+3t2+20t−3s(t)=-\frac{2 t^{3}}{3}+3 t^{2}+20 t-3s(t)=−32t3​+3t2+20t−3, t≥0t \geq 0t≥0.

Studdy Solution

STEP 1

Assumptions1. The position of the car is given by the function s(t)=−t33+3t+20t−3s(t)=-\frac{ t^{3}}{3}+3 t^{}+20 t-3s(t)=−3t3​+3t+20t−3 . t≥0t \geq0t≥0 is measured in seconds and sss is measured in meters3. We need to find the interval(s) where the car is slowing down

STEP 2

A car is slowing down when its velocity and acceleration have opposite signs. First, we need to find the velocity, which is the derivative of the position function.v(t)=s′(t)v(t) = s'(t)v(t)=s′(t)

STEP 3

Now, calculate the derivative of the position function.v(t)=−2t2+6t+20v(t) = -2t^{2} +6t +20v(t)=−2t2+6t+20

STEP 4

Next, we need to find the acceleration, which is the derivative of the velocity function.a(t)=v′(t)a(t) = v'(t)a(t)=v′(t)

STEP 5

Now, calculate the derivative of the velocity function.a(t)=−4t+a(t) = -4t +a(t)=−4t+

STEP 6

STEP 7

olve the equation v(t)=0v(t) =0v(t)=0 for ttt.−2t2+6t+20=0-2t^{2} +6t +20 =0−2t2+6t+20=0

STEP 8

olve the equation a(t)=0a(t) =0a(t)=0 for ttt.−4t+6=0-4t +6 =0−4t+6=0

STEP 9

The solutions to these equations will give us the critical points, which divide the time axis into intervals. We can then test these intervals to see where the velocity and acceleration have opposite signs.

STEP 10

Now, we need to test the intervals between the critical points to see where the car is slowing down. We do this by choosing a test point in each interval and evaluating the sign of the velocity and acceleration at these points.

STEP 11

If the signs of the velocity and acceleration are opposite at a test point, then the car is slowing down on that interval. Write these intervals in interval notation.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the interval(s) where the car slows down given $s(t)=-\frac{2 t^{3}}{3}+3 t^{2}+20 t-3$ , $t \geq 0$ .

Assumptions1. The position of the car is given by the function $s(t)=-\frac{ t^{3}}{3}+3 t^{}+20 t-3$ . $t \geq0$ is measured in seconds and $s$ is measured in meters3. We need to find the interval(s) where the car is slowing down

A car is slowing down when its velocity and acceleration have opposite signs. First, we need to find the velocity, which is the derivative of the position function.
$v(t) = s'(t)$

Now, calculate the derivative of the position function.
$v(t) = -2t^{2} +6t +20$

Next, we need to find the acceleration, which is the derivative of the velocity function.
$a(t) = v'(t)$

Now, calculate the derivative of the velocity function.
$a(t) = -4t +$

olve the equation $v(t) =0$ for $t$ .
$-2t^{2} +6t +20 =0$

olve the equation $a(t) =0$ for $t$ .
$-4t +6 =0$