QuestionFind the acceleration of the particle at $t=3$ seconds for the function $s(t)=t^{3}-3 t^{2}-8 t+1$ . Answer in $\mathrm{ft} / \mathrm{s}^{2}$ .

Studdy Solution

STEP 1

Assumptions1. The particle's position is given by the function $s(t)=t^{3}-3 t^{}-8 t+1$ . We are asked to find the acceleration at $t=3$ seconds3. The acceleration of the particle is the second derivative of the position function

STEP 2

First, we need to find the velocity of the particle, which is the first derivative of the position function.
$v(t) = s'(t)$

STEP 3

Now, calculate the first derivative of the position function.
$v(t) = s'(t) =3t^{2} -6t -8$

STEP 4

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

STEP 5

Now, calculate the second derivative of the position function, or the first derivative of the velocity function.
$a(t) = v'(t) =t -$

STEP 6

Now that we have the acceleration function, we can substitute $t=3$ seconds into the function to find the acceleration at that time.
$a(3) =6(3) -6$

STEP 7

Calculate the acceleration at $t=3$ seconds.
$a(3) =6(3) -6 =12 \, \mathrm{ft/s}^{2}$ The acceleration of the particle at $t=3$ seconds is $12 \, \mathrm{ft/s}^{2}$ .

Was this helpful?

QuestionFind the acceleration of the particle at $t=3$ seconds for the function $s(t)=t^{3}-3 t^{2}-8 t+1$ . Answer in $\mathrm{ft} / \mathrm{s}^{2}$ .

Studdy Solution

STEP 1

Assumptions1. The particle's position is given by the function $s(t)=t^{3}-3 t^{}-8 t+1$ . We are asked to find the acceleration at $t=3$ seconds3. The acceleration of the particle is the second derivative of the position function

STEP 2

First, we need to find the velocity of the particle, which is the first derivative of the position function.
$v(t) = s'(t)$

STEP 3

Now, calculate the first derivative of the position function.
$v(t) = s'(t) =3t^{2} -6t -8$

STEP 4

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

STEP 5

Now, calculate the second derivative of the position function, or the first derivative of the velocity function.
$a(t) = v'(t) =t -$

STEP 6

Now that we have the acceleration function, we can substitute $t=3$ seconds into the function to find the acceleration at that time.
$a(3) =6(3) -6$

STEP 7

Calculate the acceleration at $t=3$ seconds.
$a(3) =6(3) -6 =12 \, \mathrm{ft/s}^{2}$ The acceleration of the particle at $t=3$ seconds is $12 \, \mathrm{ft/s}^{2}$ .

Get Studdy

Studdy solves anything!

Start learning now

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

QuestionFind the acceleration of the particle at t=3t=3t=3 seconds for the function s(t)=t3−3t2−8t+1s(t)=t^{3}-3 t^{2}-8 t+1s(t)=t3−3t2−8t+1. Answer in ft/s2\mathrm{ft} / \mathrm{s}^{2}ft/s2.

Studdy Solution

STEP 1

Assumptions1. The particle's position is given by the function s(t)=t3−3t−8t+1s(t)=t^{3}-3 t^{}-8 t+1s(t)=t3−3t−8t+1 . We are asked to find the acceleration at t=3t=3t=3 seconds3. The acceleration of the particle is the second derivative of the position function

STEP 2

First, we need to find the velocity of the particle, which is the first derivative of the position function.v(t)=s′(t)v(t) = s'(t)v(t)=s′(t)

STEP 3

Now, calculate the first derivative of the position function.v(t)=s′(t)=3t2−6t−8v(t) = s'(t) =3t^{2} -6t -8v(t)=s′(t)=3t2−6t−8

STEP 4

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

STEP 5

Now, calculate the second derivative of the position function, or the first derivative of the velocity function.a(t)=v′(t)=t−a(t) = v'(t) =t -a(t)=v′(t)=t−

STEP 6

Now that we have the acceleration function, we can substitute t=3t=3t=3 seconds into the function to find the acceleration at that time.a(3)=6(3)−6a(3) =6(3) -6a(3)=6(3)−6

STEP 7

Calculate the acceleration at t=3t=3t=3 seconds.a(3)=6(3)−6=12 ft/s2a(3) =6(3) -6 =12 \, \mathrm{ft/s}^{2}a(3)=6(3)−6=12ft/s2The acceleration of the particle at t=3t=3t=3 seconds is 12 ft/s212 \, \mathrm{ft/s}^{2}12ft/s2.

Get Studdy

Studdy solves anything!

Start learning now

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

QuestionFind the acceleration of the particle at $t=3$ seconds for the function $s(t)=t^{3}-3 t^{2}-8 t+1$ . Answer in $\mathrm{ft} / \mathrm{s}^{2}$ .

Assumptions1. The particle's position is given by the function $s(t)=t^{3}-3 t^{}-8 t+1$ . We are asked to find the acceleration at $t=3$ seconds3. The acceleration of the particle is the second derivative of the position function

First, we need to find the velocity of the particle, which is the first derivative of the position function.
$v(t) = s'(t)$

Now, calculate the first derivative of the position function.
$v(t) = s'(t) =3t^{2} -6t -8$

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

Now, calculate the second derivative of the position function, or the first derivative of the velocity function.
$a(t) = v'(t) =t -$

Now that we have the acceleration function, we can substitute $t=3$ seconds into the function to find the acceleration at that time.
$a(3) =6(3) -6$

Calculate the acceleration at $t=3$ seconds.
$a(3) =6(3) -6 =12 \, \mathrm{ft/s}^{2}$ The acceleration of the particle at $t=3$ seconds is $12 \, \mathrm{ft/s}^{2}$ .