QuestionDetermine the interval(s) where the particle defined by $s(t)=\frac{2 t^{3}}{3}-10 t^{2}+48 t$ is slowing down.

Studdy Solution

STEP 1

Assumptions1. The particle's position is given by the function $s(t)=\frac{ t^{3}}{3}-10 t^{}+48 t$ where $t \geq0$ is measured in seconds and $s$ is measured in feet. . We are asked to find the interval(s) where the particle is slowing down.

STEP 2

To find where the particle is slowing down, we first need to find the velocity of the particle. The velocity is the derivative of the position function.
$v(t) = s'(t)$

STEP 3

Now, calculate the derivative of the position function to find the velocity function.
$v(t) = s'(t) = \frac{2}{3} \cdot3t^{2} -20t +48$

STEP 4

implify the velocity function.
$v(t) =2t^{2} -20t +48$

STEP 5

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

STEP 6

Now, calculate the derivative of the velocity function to find the acceleration function.
$a(t) = v'(t) =4t -20$

STEP 7

The particle is slowing down when the velocity and acceleration have opposite signs. This happens when the velocity is positive and the acceleration is negative, or when the velocity is negative and the acceleration is positive.

STEP 8

To find these intervals, we first need to find the critical points of the velocity and acceleration functions. Critical points occur where the function is zero or undefined.

STEP 9

Set the velocity function equal to zero and solve for $t$ .
$2t^{2} -20t +48 =$

STEP 10

Factor the equation to solve for $t$ .
$2(t^{2} -10t +24) =0$ $2(t -4)(t -6) =0$

STEP 11

Set each factor equal to zero and solve for $t$ .
$t -4 =0 \Rightarrow t =4$ $t -6 =0 \Rightarrow t =6$

STEP 12

The critical points of the velocity function are $t =4$ and $t =6$ .

STEP 13

Now, set the acceleration function equal to zero and solve for $t$ .
$t -20 =0$

STEP 14

olve the equation for $t$ .
$t =$

STEP 15

The critical point of the acceleration function is $t =5$ .

STEP 16

Now, we have the critical points $t =4$ , $t =5$ , and $t =6$ . We can use these to divide the number line into four intervals $[0,4)$ , $(4,5)$ , $(5,6)$ , and $(6, \infty)$ .

STEP 17

We need to test each interval to see where the velocity and acceleration have opposite signs.

STEP 18

Test the interval $[0,4)$ by choosing a test point, say $t =2$ , and substituting it into the velocity and acceleration functions.
$v(2) =2(2)^2 -20(2) +48 =8 -40 +48 =16 >0$ $a(2) =4(2) -20 =8 -20 = -12 <0$

STEP 19

Since $v() >$ and $a() <$ , the particle is slowing down on the interval $[,4)$ .

STEP 20

Test the interval $(4,5)$ by choosing a test point, say $t =4.5$ , and substituting it into the velocity and acceleration functions.
$v(4.5) =(4.5)^ -20(4.5) +48 =40.5 -90 +48 = -.5 <0$ $a(4.5) =4(4.5) -20 =18 -20 = - <0$

STEP 21

Since $v(4.5) <0$ and $a(4.5) <0$ , the particle is not slowing down on the interval $(4,5)$ .

STEP 22

Test the interval $(5,6)$ by choosing a test point, say $t =5.5$ , and substituting it into the velocity and acceleration functions.
$v(5.5) =(5.5)^ -20(5.5) +48 =60.5 -110 +48 = -1.5 <0$ $a(5.5) =4(5.5) -20 =22 -20 = >0$

STEP 23

Since $v(5.5) <0$ and $a(5.5) >0$ , the particle is slowing down on the interval $(5,6)$ .

STEP 24

Test the interval $(6, \infty)$ by choosing a test point, say $t =7$ , and substituting it into the velocity and acceleration functions.
$v(7) =(7)^ -20(7) +48 =98 -140 +48 =6 >0$ $a(7) =4(7) -20 =28 -20 =8 >0$

STEP 25

Since $v(7) >0$ and $a(7) >0$ , the particle is not slowing down on the interval $(, \infty)$ .

STEP 26

Therefore, the particle is slowing down on the intervals $[0,4)$ and $(5,6)$ .
The solution is $[0,4) \cup (5,6)$ .

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

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

Parents Influencer program Contact Policy Terms