The position of a particle at the time t > 0 is represented by $s(t) = frac 13t^3- frac 72t^2+20t-10$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Position of a particle moving along the $x$-axis questionA particle moves on a horizontal line so that its coordinate at time $t$ is $x=ln(1+2t)-t^2+2, tgeq 0.$Higher Order Derivatives problem involving the position of a particleA particle moving in a straight line has an acceleration given by $a(t)=2t$. The initial velocity of the particle is $ 2 $ cm/sec.Calculate the average acceleration and average speed of a particleSpeeding Up and Slowing Down of ParticleA particle moves along the x-axis find t when acceleration of the particle equals 0A Particle Travelling in a Straight LineFind the velocity and position of the particle as function of t?A particles moves on a horizontal line so that its coordinate at time $t$ is $x = ln (1 + 2t) − t^2 + 2, t ≥ 0.$
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The position of a particle at the time t > 0 is represented by $s(t) = frac 13t^3- frac 72t^2+20t-10$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Position of a particle moving along the $x$-axis questionA particle moves on a horizontal line so that its coordinate at time $t$ is $x=ln(1+2t)-t^2+2, tgeq 0.$Higher Order Derivatives problem involving the position of a particleA particle moving in a straight line has an acceleration given by $a(t)=2t$. The initial velocity of the particle is $ 2 $ cm/sec.Calculate the average acceleration and average speed of a particleSpeeding Up and Slowing Down of ParticleA particle moves along the x-axis find t when acceleration of the particle equals 0A Particle Travelling in a Straight LineFind the velocity and position of the particle as function of t?A particles moves on a horizontal line so that its coordinate at time $t$ is $x = ln (1 + 2t) − t^2 + 2, t ≥ 0.$
$begingroup$
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
asked Apr 1 at 7:26
Aman KhanAman Khan
326
$endgroup$
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
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
asked Apr 1 at 7:26
Aman KhanAman Khan
326
$endgroup$
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
calculus derivatives
asked Apr 1 at 7:26
Aman KhanAman Khan
326
asked Apr 1 at 7:26
Aman KhanAman Khan
326
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
326
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
4 Answers
4
active
oldest
votes
$begingroup$
$s''(t) >0$ if $t >frac 7 2$ so it is speending up for $t >frac 7 2$.
answered Apr 1 at 7:29
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
$endgroup$
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.
answered Apr 1 at 7:31
Eevee TrainerEevee Trainer
10.5k31842
$endgroup$
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 $s'' > 0$
$s'' = 2t - 7 > 0\
t > frac 72$
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 |
$begingroup$
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.
answered Apr 1 at 7:39
Yasir MKYasir MK
114
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
$s''(t) >0$ if $t >frac 7 2$ so it is speending up for $t >frac 7 2$.
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 $t >frac 7 2$.
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 $t >frac 7 2$.
answered Apr 1 at 7:29
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
$endgroup$
$s''(t) >0$ if $t >frac 7 2$ so it is speending up for $t >frac 7 2$.
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
answered Apr 1 at 7:29
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
75.3k53270
add a comment |
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.
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.
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.
answered Apr 1 at 7:31
Eevee TrainerEevee Trainer
10.5k31842
$endgroup$
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.
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
answered Apr 1 at 7:31
Eevee TrainerEevee Trainer
10.5k31842
10.5k31842
add a comment |
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 $s'' > 0$
$s'' = 2t - 7 > 0\
t > frac 72$
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 $s'' > 0$
$s'' = 2t - 7 > 0\
t > frac 72$
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 $s'' > 0$
$s'' = 2t - 7 > 0\
t > frac 72$
answered Apr 1 at 7:35
Doug MDoug M
1,996412
$endgroup$
"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 $s'' > 0$
$s'' = 2t - 7 > 0\
t > frac 72$
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
answered Apr 1 at 7:35
Doug MDoug M
1,996412
1,996412
$begingroup$
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 |
$begingroup$
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
$begingroup$
Problem specifically mentions "speeding up", which means increase of speed or acceleration.
$endgroup$
– Aman Khan
Apr 1 at 8:22
$begingroup$
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
$begingroup$
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 |
$begingroup$
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.
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.
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.
answered Apr 1 at 7:39
Yasir MKYasir MK
114
$endgroup$
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.
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
answered Apr 1 at 7:39
Yasir MKYasir MK
114
114
add a comment |
add a comment |
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$begingroup$
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