Equation of free fall

prdk21

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Jan 9, 2019
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so here is the question:

"The equations for free fall at the surfaces of Pluto is s = 5.28t^2;

where s is in meters, and t is in seconds. How long would it

take an object to fall from rest to reach a velocity of 36 km/h?"

For some reason this question is really stumping me. I don't know where to start, and I don't even really understand what the
equations represent. Can someone help me understand this conceptually and
give me some steps to follow? Thanks!
 

Subhotosh Khan

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18,243
so here is the question:

"The equations for free fall at the surfaces of Pluto is s = 5.28t^2;

where s is in meters, and t is in seconds. How long would it

take an object to fall from rest to reach a velocity of 36 km/h?"

For some reason this question is really stumping me. I don't know where to start, and I don't even really understand what the
equations represent. Can someone help me understand this conceptually and
give me some steps to follow? Thanks!
You know how to calculate derivatives - correct?

Do you know the relation between speed and distance travelled?
 

topsquark

Full Member
Joined
Aug 27, 2012
Messages
332
so here is the question:

"The equations for free fall at the surfaces of Pluto is s = 5.28t^2;

where s is in meters, and t is in seconds. How long would it

take an object to fall from rest to reach a velocity of 36 km/h?"

For some reason this question is really stumping me. I don't know where to start, and I don't even really understand what the
equations represent. Can someone help me understand this conceptually and
give me some steps to follow? Thanks!
From a Physicist's perspective: From the given equation what is a? Do you know an equation for speed in terms of the acceleration and time?

-Dan
 

Dr.Peterson

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Joined
Nov 12, 2017
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3,592
so here is the question:

"The equations for free fall at the surfaces of Pluto is s = 5.28t^2;

where s is in meters, and t is in seconds. How long would it

take an object to fall from rest to reach a velocity of 36 km/h?"

For some reason this question is really stumping me. I don't know where to start, and I don't even really understand what the
equations represent
. Can someone help me understand this conceptually and
give me some steps to follow? Thanks!
As you can see from the other answers, we'll need to know your background -- are you taking physics, or calculus, or algebra, or what? What do you know about distance, velocity, and acceleration? Do you know any relevant formulas? That is, what DO you understand, so we can use that to help you move forward?

This is covered in this summary of our submission guidelines, which you should have read. It really helps in getting a good answer quickly.
 

Jomo

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Joined
Dec 30, 2014
Messages
3,535
so here is the question:

"The equations for free fall at the surfaces of Pluto is s = 5.28t^2;

where s is in meters, and t is in seconds. How long would it

take an object to fall from rest to reach a velocity of 36 km/h?"

For some reason this question is really stumping me. I don't know where to start, and I don't even really understand what the
equations represent. Can someone help me understand this conceptually and
give me some steps to follow? Thanks!
S usually stands for distance traveled (and in this case it does) and t is time in seconds. S0 s=5.28t^2 means that if you plug in some time for t, then when you calculate this (2.58t^2) you get S in meters (meters because it said s is in meters. It could have been in feet or miles...).

Can you think of any numerical example for velocity? Remember velocity is simply speed with direction.

An example for speed is 50 miles per hour which is the same as 50 miles / hour. Velocity could be 50 miles per hour downwards. 50 miles / hour is (the change in distance)/ (the change in time). This says that velocity, v =(change in distance)/(change in time) =
(change in S)/(change in t) = (With some hand waving) dS/dt. That is V = dS/dt

So we want to know what t equals when v= 36km/hr.

Simply find dS/dt, which will be in terms of t, and set it equal to 36km/h and then solve for t.

Let us know how you make out.
 

prdk21

New member
Joined
Jan 9, 2019
Messages
9
You know how to calculate derivatives - correct?

Do you know the relation between speed and distance travelled?
HI,
I do know how to calculate derivatives. I guess why this is giving me so much trouble
is because I thought normally displacement equations have a term that is cubed.
I guess that is not always the case? I think I may have been overthinking
this and not realized that all I had to do was derive the equation with respect
to time.
 

prdk21

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Joined
Jan 9, 2019
Messages
9
As you can see from the other answers, we'll need to know your background -- are you taking physics, or calculus, or algebra, or what? What do you know about distance, velocity, and acceleration? Do you know any relevant formulas? That is, what DO you understand, so we can use that to help you move forward?

This is covered in this summary of our submission guidelines, which you should have read. It really helps in getting a good answer quickly.
Hi,

Sorry for the lack of background. I am currently taking a graduate level dynamic
meteorology course. It is calculus heavy, but I am assuming this is more physics
based - if that makes sense.
I understand that velocity is the derivative of displacement, and accleration
is the derivative of velocity.
I guess my biggest confusion is the wording and why the displacement equation
is only squared when most I have seen have a cubed term.
Thanks for your response!
 

prdk21

New member
Joined
Jan 9, 2019
Messages
9
From a Physicist's perspective: From the given equation what is a? Do you know an equation for speed in terms of the acceleration and time?

-Dan
Hi,
That was the only information I was given. That was another point of confusion
for me, because I am used to doing the standard free fall equation including some
form of at (acceleration times time) aspect. And clearly I can't assume that
acceleration is g since it is on a different planet.
Thank you for your response!
 

prdk21

New member
Joined
Jan 9, 2019
Messages
9
S usually stands for distance traveled (and in this case it does) and t is time in seconds. S0 s=5.28t^2 means that if you plug in some time for t, then when you calculate this (2.58t^2) you get S in meters (meters because it said s is in meters. It could have been in feet or miles...).

Can you think of any numerical example for velocity? Remember velocity is simply speed with direction.

An example for speed is 50 miles per hour which is the same as 50 miles / hour. Velocity could be 50 miles per hour downwards. 50 miles / hour is (the change in distance)/ (the change in time). This says that velocity, v =(change in distance)/(change in time) =
(change in S)/(change in t) = (With some hand waving) dS/dt. That is V = dS/dt

So we want to know what t equals when v= 36km/hr.

Simply find dS/dt, which will be in terms of t, and set it equal to 36km/h and then solve for t.

Let us know how you make out.
Hi,
Thank you for your detailed response!
This was my initial assumption, which I should have made clear. I guess it seemed
a little too simple and I got confused because when I looked at all other examples
displacement had a cubed term so I thought there was something I was missing.
Then I looked at the equation and realized none of the units balanced out which
also really freaked me out. But I guess I was just over thinking!
Thanks you so much!
 

prdk21

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Joined
Jan 9, 2019
Messages
9
Response

Hi all,

I apologize for my lack of information I put in my thread.
I am currently a grad student in applied meteorology and this class is for
Meteorology Dynamics. It is heavily calculus/physics based.
This is my first post, so I am not sure if this response will notify you guys or
not.
Anyways, I do understand derivatives. I get that velocity is the derivative of
displacement, and that was my initial thought when I saw this problem.
The two things that tripped me up were the units (they aren't equal on each side
of the equation, so my chemistry background had a panic attack) and the fact
that the displacement equation was only dependent on time squared. I thought
normally there was a cubed term and some acceleration (usually gravity) term as
well. But maybe I am just over complicating things.
Thank you for your time and responses!
-Danny
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,243
so here is the question:

"The equations for free fall at the surfaces of Pluto is s = 5.28t^2;

where s is in meters, and t is in seconds. How long would it

take an object to fall from rest to reach a velocity of 36 km/h?"

For some reason this question is really stumping me. I don't know where to start, and I don't even really understand what the
equations represent. Can someone help me understand this conceptually and
give me some steps to follow? Thanks!
When you differentiate the "s" twice - you'll get a constant term.

That "constant acceleration" makes your problem a special case - where we can use the following equation:

constant acceleration = (velocityfinal - velocityinital)/(lapsed time)

continue.....
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
3,535
The two things that tripped me up were the units (they aren't equal on each side
of the equation, so my chemistry background had a panic attack) and the fact
that the displacement equation was only dependent on time squared.
5.28 should really be 5.28m/s^2. So in the end, when you multiply (5.28m/s^2)(?s^2) the units will be meters which is fine.

Suppose you have a constant acceleration, a

So a=a
Then v = v0 + atThen S= S0 + .5at2
I simply integrated to get the next equation.

It seems as though your S0 = 0 (after all S(0)=0) and your a is 10.56m/s2 . The units really do work out.
 
Last edited:

topsquark

Full Member
Joined
Aug 27, 2012
Messages
332
I guess my biggest confusion is the wording and why the displacement equation
is only squared when most I have seen have a cubed term.
Thanks for your response!
Most of the time in Introductory Physics we use the constant acceleration equations, even if it's a Calculus based class. The displacement equation is given as a series of some power of time times a constant, which have various meanings.
\(\displaystyle s = s_0 + v_0t + \dfrac{1}{2}at^2 + \dfrac{1}{6}jt^3 \text{ ...}\)

Here, the s stands for displacement, v for velocity, a for acceleration, and j for "jerk." (Yes folks, this really is the name for it.) I've never heard of a problem that uses any more terms than this one. This is possibly the equation you are referring to when you say you've seen the "cubed" term.

The only other example I've seen is something of the form \(\displaystyle s = f(t)\) , where f(t) is some function of time. (Depending on your textbook some other letter than f might be used.) This is the form we usually reserve for Calculus based problems and you can get velocity, acceleration, etc. by taking derivatives.

-Dan
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
3,535
Most of the time in Introductory Physics we use the constant acceleration equations, even if it's a Calculus based class. The displacement equation is given as a series of some power of time times a constant, which have various meanings.
\(\displaystyle s = s_0 + v_0t + \dfrac{1}{2}at^2 + \dfrac{1}{6}jt^3 \text{ ...}\)

Here, the s stands for displacement, v for velocity, a for acceleration, and j for "jerk." (Yes folks, this really is the name for it.) I've never heard of a problem that uses any more terms than this one. This is possibly the equation you are referring to when you say you've seen the "cubed" term.

The only other example I've seen is something of the form \(\displaystyle s = f(t)\) , where f(t) is some function of time. (Depending on your textbook some other letter than f might be used.) This is the form we usually reserve for Calculus based problems and you can get velocity, acceleration, etc. by taking derivatives.

-Dan
Yes, I heard of jerk. It is da/dt if I remember correctly (and according to your formula). Unfortunately my Physics professor never spoke about it at all other than to mention it.

BTW, I have been meaning to say that it is great to have a Physicist helping out here. THANKS!
 
Last edited:

prdk21

New member
Joined
Jan 9, 2019
Messages
9
Most of the time in Introductory Physics we use the constant acceleration equations, even if it's a Calculus based class. The displacement equation is given as a series of some power of time times a constant, which have various meanings.
\(\displaystyle s = s_0 + v_0t + \dfrac{1}{2}at^2 + \dfrac{1}{6}jt^3 \text{ ...}\)

Here, the s stands for displacement, v for velocity, a for acceleration, and j for "jerk." (Yes folks, this really is the name for it.) I've never heard of a problem that uses any more terms than this one. This is possibly the equation you are referring to when you say you've seen the "cubed" term.

The only other example I've seen is something of the form \(\displaystyle s = f(t)\) , where f(t) is some function of time. (Depending on your textbook some other letter than f might be used.) This is the form we usually reserve for Calculus based problems and you can get velocity, acceleration, etc. by taking derivatives.

-Dan
Ahh, okay. I must be remembering that from Physics that I took quite a few years ago.
That was really helpful, thank you so much!
 
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