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The distance travelled, $s$ in meters, by an object after any number of seconds,$t$, is given by $s(t)=5t^4 - 16t + 24$ [on hold]

$S=5t^4 - 16t + 24$Up and Down Motion (Two objects meeting in time?)How do I determine the distance of an object if the acceleration is constantly changing?An object is travelling in a straight line. Its distance, s meters, from a fixed point at time t seconds is given by the expressionDistance travelled on a curvelinear path and the coordinate of pointsPre-calculus - Deriving position from accelerationThe differences between A and B are given by $d(t) = 50(e^-0.4t-e^-.08t)$, is A ever greater than B?Find both the velocity v and directed distance s after 1.7 seconds.A 5kg object is acted on by a horizontal force, find the velocity when t = 5 secondsA Particle Travelling in a Straight LineCalculus physics

-3

$begingroup$

$$s(t)=5t^4 - 16t + 24$$

FIND:

a) find the speed of the object after $3$ seconds

b) the acceleration after $10$ seconds

c) after how many seconds does the speed equal to zero

d) the distance travelled when the object is stationary

calculus

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edited Mar 11 at 20:05

dantopa

6,64942245

asked Mar 11 at 19:55

Ray MarmionRay Marmion

1

New contributor

Ray Marmion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by John Omielan, Lord Shark the Unknown, mrtaurho, José Carlos Santos, Cesareo Mar 12 at 9:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Omielan, Lord Shark the Unknown, mrtaurho, José Carlos Santos, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  We're not doing your homework. Hint: how do you find speed given distance?
  $endgroup$
  – Sean Roberson
  Mar 11 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wouldn't of put the question on the forum if I knew how to do it.....
  $endgroup$
  – Ray Marmion
  Mar 11 at 20:03
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
  $endgroup$
  – dantopa
  Mar 11 at 20:04
  

  
  
  
  

add a comment | 

-3

$begingroup$

$$s(t)=5t^4 - 16t + 24$$

FIND:

a) find the speed of the object after $3$ seconds

b) the acceleration after $10$ seconds

c) after how many seconds does the speed equal to zero

d) the distance travelled when the object is stationary

calculus

share|cite|improve this question

edited Mar 11 at 20:05

dantopa

6,64942245

asked Mar 11 at 19:55

Ray MarmionRay Marmion

1

New contributor

Ray Marmion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by John Omielan, Lord Shark the Unknown, mrtaurho, José Carlos Santos, Cesareo Mar 12 at 9:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Omielan, Lord Shark the Unknown, mrtaurho, José Carlos Santos, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  We're not doing your homework. Hint: how do you find speed given distance?
  $endgroup$
  – Sean Roberson
  Mar 11 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wouldn't of put the question on the forum if I knew how to do it.....
  $endgroup$
  – Ray Marmion
  Mar 11 at 20:03
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
  $endgroup$
  – dantopa
  Mar 11 at 20:04
  

  
  
  
  

add a comment | 

$begingroup$

$$s(t)=5t^4 - 16t + 24$$

FIND:

a) find the speed of the object after $3$ seconds

b) the acceleration after $10$ seconds

c) after how many seconds does the speed equal to zero

d) the distance travelled when the object is stationary

calculus

share|cite|improve this question

edited Mar 11 at 20:05

dantopa

6,64942245

asked Mar 11 at 19:55

Ray MarmionRay Marmion

1

New contributor

Ray Marmion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 11 at 20:05

dantopa

6,64942245

asked Mar 11 at 19:55

Ray MarmionRay Marmion

1

New contributor

Ray Marmion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 11 at 20:05

dantopa

6,64942245

asked Mar 11 at 19:55

Ray MarmionRay Marmion

1

New contributor

Ray Marmion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 11 at 20:05

dantopa

6,64942245

edited Mar 11 at 20:05

dantopa

6,64942245

asked Mar 11 at 19:55

Ray MarmionRay Marmion

1

New contributor

Ray Marmion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 11 at 19:55

Ray MarmionRay Marmion

1

1 Answer
1

active

oldest

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0

$begingroup$

For part a, differentiate $s(t)$. Remember, $v(t)=s'(t)$. Taking the derivative gives you $v(t)=20t^3-16$. (Just basic Power Rule stuff). Substituting in gives $v(3)=20(27)-16=524.$

Part (b): If you want to find the acceleration, find $s''(t)=v'(t)$, as the acceleration is defined as derivative of velocity. Differentiating gives $a(t)=60t^2$, so $a(10)=6000$.

Part (c): This part is a bit more involved, but doable. The particle stops when $v(t)=0$. Just set $v(t)=20t^3-16=0$ and solve. You can factor out $4$, so you get $4(5t^3-4)=0$, so $t=sqrt[3]frac45=fracsqrt[3]1005$.

Part(d): The total distance traveled is just $s(fracsqrt[3]1005)-s(0)$. You can just evaluate the function at those points. That you should be able to do, as it requires zero calculus.

Happy mathing!

share|cite|improve this answer

answered Mar 12 at 2:00

A RA R

766

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This should be basic Power Rule stuff. Why not try it first so that we can help you better?
  $endgroup$
  – A R
  Mar 12 at 2:04
  

  
  
  
  

add a comment | 

0

$begingroup$

For part a, differentiate $s(t)$. Remember, $v(t)=s'(t)$. Taking the derivative gives you $v(t)=20t^3-16$. (Just basic Power Rule stuff). Substituting in gives $v(3)=20(27)-16=524.$

Part (b): If you want to find the acceleration, find $s''(t)=v'(t)$, as the acceleration is defined as derivative of velocity. Differentiating gives $a(t)=60t^2$, so $a(10)=6000$.

Part (c): This part is a bit more involved, but doable. The particle stops when $v(t)=0$. Just set $v(t)=20t^3-16=0$ and solve. You can factor out $4$, so you get $4(5t^3-4)=0$, so $t=sqrt[3]frac45=fracsqrt[3]1005$.

Part(d): The total distance traveled is just $s(fracsqrt[3]1005)-s(0)$. You can just evaluate the function at those points. That you should be able to do, as it requires zero calculus.

Happy mathing!

share|cite|improve this answer

answered Mar 12 at 2:00

A RA R

766

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This should be basic Power Rule stuff. Why not try it first so that we can help you better?
  $endgroup$
  – A R
  Mar 12 at 2:04
  

  
  
  
  

add a comment | 

0

$begingroup$

For part a, differentiate $s(t)$. Remember, $v(t)=s'(t)$. Taking the derivative gives you $v(t)=20t^3-16$. (Just basic Power Rule stuff). Substituting in gives $v(3)=20(27)-16=524.$

Part (b): If you want to find the acceleration, find $s''(t)=v'(t)$, as the acceleration is defined as derivative of velocity. Differentiating gives $a(t)=60t^2$, so $a(10)=6000$.

Part (c): This part is a bit more involved, but doable. The particle stops when $v(t)=0$. Just set $v(t)=20t^3-16=0$ and solve. You can factor out $4$, so you get $4(5t^3-4)=0$, so $t=sqrt[3]frac45=fracsqrt[3]1005$.

Part(d): The total distance traveled is just $s(fracsqrt[3]1005)-s(0)$. You can just evaluate the function at those points. That you should be able to do, as it requires zero calculus.

Happy mathing!

share|cite|improve this answer

answered Mar 12 at 2:00

A RA R

766

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This should be basic Power Rule stuff. Why not try it first so that we can help you better?
  $endgroup$
  – A R
  Mar 12 at 2:04
  

  
  
  
  

add a comment | 

$begingroup$

For part a, differentiate $s(t)$. Remember, $v(t)=s'(t)$. Taking the derivative gives you $v(t)=20t^3-16$. (Just basic Power Rule stuff). Substituting in gives $v(3)=20(27)-16=524.$

Part (b): If you want to find the acceleration, find $s''(t)=v'(t)$, as the acceleration is defined as derivative of velocity. Differentiating gives $a(t)=60t^2$, so $a(10)=6000$.

Part (c): This part is a bit more involved, but doable. The particle stops when $v(t)=0$. Just set $v(t)=20t^3-16=0$ and solve. You can factor out $4$, so you get $4(5t^3-4)=0$, so $t=sqrt[3]frac45=fracsqrt[3]1005$.

Part(d): The total distance traveled is just $s(fracsqrt[3]1005)-s(0)$. You can just evaluate the function at those points. That you should be able to do, as it requires zero calculus.

Happy mathing!

share|cite|improve this answer

answered Mar 12 at 2:00

A RA R

766

$endgroup$

share|cite|improve this answer

answered Mar 12 at 2:00

A RA R

766

answered Mar 12 at 2:00

A RA R

766

answered Mar 12 at 2:00

A RA R

766