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The position of a particle at the time t > 0 is represented by $s(t) = frac 13t^3- frac 72t^2+20t-10$

0

$begingroup$

At which time particle is speeding up?

From my experience I think velocity is calculated by taking first derivative s'(t), which is $s'(t)=t^2-7t+20$ and acceleration calculated by taking second derivative S''(t) which is $s''(t)=2t-7$.

How to find at which time it speeding up?

calculus derivatives

share|cite|improve this question

edited Apr 1 at 7:29

Aman Khan

asked Apr 1 at 7:26

Aman KhanAman Khan

326

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Be careful with the "speeding up". In general, a particle with $s''(t_0)> 0$ is really "slowing down" at time $t_0$ if $s'(t_0) < 0$. (Because the speed is actually decreasing at this point!) If "speeding up" means the speed (not just velocity!) is increasing, you want the times $t$ such that $s''(t)$ and $s'(t)$ have the same sign, rather than just times where $s''(t) > 0$.
  $endgroup$
  – Minus One-Twelfth
  Apr 1 at 8:22
  
  

  
  
  
  

add a comment | 

0

$begingroup$

At which time particle is speeding up?

From my experience I think velocity is calculated by taking first derivative s'(t), which is $s'(t)=t^2-7t+20$ and acceleration calculated by taking second derivative S''(t) which is $s''(t)=2t-7$.

How to find at which time it speeding up?

calculus derivatives

share|cite|improve this question

edited Apr 1 at 7:29

Aman Khan

asked Apr 1 at 7:26

Aman KhanAman Khan

326

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Be careful with the "speeding up". In general, a particle with $s''(t_0)> 0$ is really "slowing down" at time $t_0$ if $s'(t_0) < 0$. (Because the speed is actually decreasing at this point!) If "speeding up" means the speed (not just velocity!) is increasing, you want the times $t$ such that $s''(t)$ and $s'(t)$ have the same sign, rather than just times where $s''(t) > 0$.
  $endgroup$
  – Minus One-Twelfth
  Apr 1 at 8:22
  
  

  
  
  
  

add a comment | 

$begingroup$

At which time particle is speeding up?

From my experience I think velocity is calculated by taking first derivative s'(t), which is $s'(t)=t^2-7t+20$ and acceleration calculated by taking second derivative S''(t) which is $s''(t)=2t-7$.

How to find at which time it speeding up?

calculus derivatives

share|cite|improve this question

edited Apr 1 at 7:29

Aman Khan

asked Apr 1 at 7:26

Aman KhanAman Khan

326

$endgroup$

share|cite|improve this question

edited Apr 1 at 7:29

Aman Khan

asked Apr 1 at 7:26

Aman KhanAman Khan

326

share|cite|improve this question

edited Apr 1 at 7:29

Aman Khan

asked Apr 1 at 7:26

Aman KhanAman Khan

326

asked Apr 1 at 7:26

Aman KhanAman Khan

326

asked Apr 1 at 7:26

Aman KhanAman Khan

326

4 Answers
4

active

oldest

votes

2

$begingroup$

$s''(t) >0$ if $t >frac 7 2$ so it is speending up for $t >frac 7 2$.

share|cite|improve this answer

answered Apr 1 at 7:29

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

$endgroup$

add a comment | 

1

$begingroup$

If I understand correctly, "speeding up" refers to positive acceleration. Therefore, you want to find the second derivative of $s$, i.e. $s''(t)$, and find for which values of $t$ we have $s''(t) > 0$.

This is best approached by considering the roots of the equation, and the behavior on other side of the roots. Since $s$ is a cubic, $s''$ will be linear, so one side of its root (of which there is only one) it will be negative (slowing down) and on the other $s''$ will be positive (speeding up).

Determine that and you will have an infinite interval, from that root, to the corresponding infinity, on which the particle speeds up.

share|cite|improve this answer

answered Apr 1 at 7:31

Eevee TrainerEevee Trainer

10.5k31842

$endgroup$

add a comment | 

1

$begingroup$

"speed" does not have direction; velocity does.

$textvelocity = s'\
text speed = |s'|$

$s' = t^2 - 7t + 20$

$s' > 0 $ for all $t.$

That makes things easier. Then we only need to know when $s'' > 0$

$s'' = 2t - 7 > 0\
t > frac 72$

share|cite|improve this answer

answered Apr 1 at 7:35

Doug MDoug M

1,996412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Problem specifically mentions "speeding up", which means increase of speed or acceleration.
  $endgroup$
  – Aman Khan
  Apr 1 at 8:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you are traveling to the right, and your acceleration is positive, you are speeding up. If you are traveling to the left, and your acceleration is positive, you are slowing down. If you are traveling to the left and your acceleration is negative, you are slowing down.
  $endgroup$
  – Doug M
  Apr 2 at 18:19
  

  
  
  
  

add a comment | 

1

$begingroup$

Equate this equation to $zero $ to find the time it starts speeding up in other words"accelarating"
$s′′(t)=2t−7$
$2t−7=0$
$2t=7$
$t=frac 72$ , that's the time it speeds up.

share|cite|improve this answer

answered Apr 1 at 7:39

Yasir MKYasir MK

114

$endgroup$

add a comment | 

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2

$begingroup$

$s''(t) >0$ if $t >frac 7 2$ so it is speending up for $t >frac 7 2$.

share|cite|improve this answer

answered Apr 1 at 7:29

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

$endgroup$

add a comment | 

2

$begingroup$

$s''(t) >0$ if $t >frac 7 2$ so it is speending up for $t >frac 7 2$.

share|cite|improve this answer

answered Apr 1 at 7:29

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

$endgroup$

add a comment | 

$begingroup$

$s''(t) >0$ if $t >frac 7 2$ so it is speending up for $t >frac 7 2$.

share|cite|improve this answer

answered Apr 1 at 7:29

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

$endgroup$

share|cite|improve this answer

answered Apr 1 at 7:29

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

answered Apr 1 at 7:29

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

answered Apr 1 at 7:29

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

1

$begingroup$

If I understand correctly, "speeding up" refers to positive acceleration. Therefore, you want to find the second derivative of $s$, i.e. $s''(t)$, and find for which values of $t$ we have $s''(t) > 0$.

This is best approached by considering the roots of the equation, and the behavior on other side of the roots. Since $s$ is a cubic, $s''$ will be linear, so one side of its root (of which there is only one) it will be negative (slowing down) and on the other $s''$ will be positive (speeding up).

Determine that and you will have an infinite interval, from that root, to the corresponding infinity, on which the particle speeds up.

share|cite|improve this answer

answered Apr 1 at 7:31

Eevee TrainerEevee Trainer

10.5k31842

$endgroup$

add a comment | 

1

$begingroup$

If I understand correctly, "speeding up" refers to positive acceleration. Therefore, you want to find the second derivative of $s$, i.e. $s''(t)$, and find for which values of $t$ we have $s''(t) > 0$.

This is best approached by considering the roots of the equation, and the behavior on other side of the roots. Since $s$ is a cubic, $s''$ will be linear, so one side of its root (of which there is only one) it will be negative (slowing down) and on the other $s''$ will be positive (speeding up).

Determine that and you will have an infinite interval, from that root, to the corresponding infinity, on which the particle speeds up.

share|cite|improve this answer

answered Apr 1 at 7:31

Eevee TrainerEevee Trainer

10.5k31842

$endgroup$

add a comment | 

$begingroup$

If I understand correctly, "speeding up" refers to positive acceleration. Therefore, you want to find the second derivative of $s$, i.e. $s''(t)$, and find for which values of $t$ we have $s''(t) > 0$.

This is best approached by considering the roots of the equation, and the behavior on other side of the roots. Since $s$ is a cubic, $s''$ will be linear, so one side of its root (of which there is only one) it will be negative (slowing down) and on the other $s''$ will be positive (speeding up).

Determine that and you will have an infinite interval, from that root, to the corresponding infinity, on which the particle speeds up.

share|cite|improve this answer

answered Apr 1 at 7:31

Eevee TrainerEevee Trainer

10.5k31842

$endgroup$

share|cite|improve this answer

answered Apr 1 at 7:31

Eevee TrainerEevee Trainer

10.5k31842

answered Apr 1 at 7:31

Eevee TrainerEevee Trainer

10.5k31842

answered Apr 1 at 7:31

Eevee TrainerEevee Trainer

10.5k31842

1

$begingroup$

"speed" does not have direction; velocity does.

$textvelocity = s'\
text speed = |s'|$

$s' = t^2 - 7t + 20$

$s' > 0 $ for all $t.$

That makes things easier. Then we only need to know when $s'' > 0$

$s'' = 2t - 7 > 0\
t > frac 72$

share|cite|improve this answer

answered Apr 1 at 7:35

Doug MDoug M

1,996412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Problem specifically mentions "speeding up", which means increase of speed or acceleration.
  $endgroup$
  – Aman Khan
  Apr 1 at 8:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you are traveling to the right, and your acceleration is positive, you are speeding up. If you are traveling to the left, and your acceleration is positive, you are slowing down. If you are traveling to the left and your acceleration is negative, you are slowing down.
  $endgroup$
  – Doug M
  Apr 2 at 18:19
  

  
  
  
  

add a comment | 

1

$begingroup$

"speed" does not have direction; velocity does.

$textvelocity = s'\
text speed = |s'|$

$s' = t^2 - 7t + 20$

$s' > 0 $ for all $t.$

That makes things easier. Then we only need to know when $s'' > 0$

$s'' = 2t - 7 > 0\
t > frac 72$

share|cite|improve this answer

answered Apr 1 at 7:35

Doug MDoug M

1,996412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Problem specifically mentions "speeding up", which means increase of speed or acceleration.
  $endgroup$
  – Aman Khan
  Apr 1 at 8:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you are traveling to the right, and your acceleration is positive, you are speeding up. If you are traveling to the left, and your acceleration is positive, you are slowing down. If you are traveling to the left and your acceleration is negative, you are slowing down.
  $endgroup$
  – Doug M
  Apr 2 at 18:19
  

  
  
  
  

add a comment | 

$begingroup$

"speed" does not have direction; velocity does.

$textvelocity = s'\
text speed = |s'|$

$s' = t^2 - 7t + 20$

$s' > 0 $ for all $t.$

That makes things easier. Then we only need to know when $s'' > 0$

$s'' = 2t - 7 > 0\
t > frac 72$

share|cite|improve this answer

answered Apr 1 at 7:35

Doug MDoug M

1,996412

$endgroup$

share|cite|improve this answer

answered Apr 1 at 7:35

Doug MDoug M

1,996412

answered Apr 1 at 7:35

Doug MDoug M

1,996412

answered Apr 1 at 7:35

Doug MDoug M

1,996412

1

$begingroup$

Equate this equation to $zero $ to find the time it starts speeding up in other words"accelarating"
$s′′(t)=2t−7$
$2t−7=0$
$2t=7$
$t=frac 72$ , that's the time it speeds up.

share|cite|improve this answer

answered Apr 1 at 7:39

Yasir MKYasir MK

114

$endgroup$

add a comment | 

1

$begingroup$

Equate this equation to $zero $ to find the time it starts speeding up in other words"accelarating"
$s′′(t)=2t−7$
$2t−7=0$
$2t=7$
$t=frac 72$ , that's the time it speeds up.

share|cite|improve this answer

answered Apr 1 at 7:39

Yasir MKYasir MK

114

$endgroup$

add a comment | 

$begingroup$

Equate this equation to $zero $ to find the time it starts speeding up in other words"accelarating"
$s′′(t)=2t−7$
$2t−7=0$
$2t=7$
$t=frac 72$ , that's the time it speeds up.

share|cite|improve this answer

answered Apr 1 at 7:39

Yasir MKYasir MK

114

$endgroup$

share|cite|improve this answer

answered Apr 1 at 7:39

Yasir MKYasir MK

114

answered Apr 1 at 7:39

Yasir MKYasir MK

114

answered Apr 1 at 7:39

Yasir MKYasir MK

114