Kinematics Equation: S = Ut + ½a t^2 - Why "s"?

[parshyaa]

Why 's' in all kinematics equation is displacement.
S = Ut + ½a t^2 , v^2 = u^2 + 2aS

• is it because S in v = dS/dt is a small distance , very small distance and we treat it as a position of particle at a time t1 during Δt→0 and its direction is from t1(intial position) to t2(final position) . If not please help me.

 

[jbriggs444]

Why 's' in all kinematics equation is displacement.
S = Ut + ½a t^2 , v^2 = u^2 + 2aS

• is it because S in v = dS/dt is a small distance , very small distance and we treat it as a position of particle at a time t1 during Δt→0 and its direction is from t1(intial position) to t2(final position) . If not please help me.

It is not very clear what question you are trying to ask.

S is a displacement, not a position. If you wanted to denote a position, you would typically use "x" as the variable name instead of "s". The choice of variable name is purely based on convention. There is no deep physical significance. A displacement is a directed distance between one position and another.

When we write v = $\frac{ds}{dt}$, this means that there is a particular ratio between distance traversed in an interval and time elapsed in that interval that is approached as the interval decreases toward zero and that that ratio is v. This notation is used in Calculus where the meaning is defined formally with a quantified expression involving epsilons and deltas.

Technically, the S in v = dS/dt does not denote any particular fixed small displacement. It does not denote anything at all. It is only a part of the notation whose meaning is as above. Even so, you can think of dS/dt as denoting a very small displacement (dS) divided by a very short time interval (dt) without going badly wrong. [In the 20th century, mathematicians came up with theoretical support (non-standard analysis and the Transfer Principle) which can make it formally correct to consider dS/dt as the ratio between an infinitesimal displacement and an infinitesimal time].

 

[parshyaa]

Ok , let us say that V(av) = ΔS/Δt is the average speed of the particle which means it is for a time interval , let's say t1 and t2. But if we want to find instantaneous speed of a particle at time t1 , supose u want to find speed at t1 = 4sec, to calculate this we have only one information, I.e average speed between t1 and t2 , Vav = Δs/Δt, here Δs/Δt is the average speed between t1 and t2 what if we make it average speed between t1 and t1 which means instantaneous speed at t1, in order to do so we have to make t2 -t1 =Δt = 0 , but doing so will give us average speed undefined so we can't do this therefore we make Δt tooooo close to 0 so that we can say that finding speed between t1 =4sec is same as finding average speed at time t1 =4 and t2=4.000000000000000001 sec ≈ 4 sec. Therefore Δt becomes dt , as time and distance are related Δs also tends to 0 and becomes dt therefore we say that V = ds/dt and it tell us the speed at a instant but its a still a average speed of very very small time frame. But if we talk instantaneous speed geometrically it will be the slope of tangent at that point because geometrically we can define it as a slope of line joining position at t1 and t2 and as t2 reaches t1 slope of line becomes slope of point or slope of tangent . but I am not clear why and how S is displacement , is it because accelaration is a vector.

It is not very clear what question you are trying to ask.

S is a displacement, not a position. If you wanted to denote a position, you would typically use "x" as the variable name instead of "s". The choice of variable name is purely based on convention. There is no deep physical significance. A displacement is a directed distance between one position and another.

When we write v = $\frac{ds}{dt}$, this means that there is a particular ratio between distance traversed in an interval and time elapsed in that interval that is approached as the interval decreases toward zero and that that ratio is v. This notation is used in Calculus where the meaning is defined formally with a quantified expression involving epsilons and deltas.

Technically, the S in v = dS/dt does not denote any particular fixed small displacement. It does not denote anything at all. It is only a part of the notation whose meaning is as above. Even so, you can think of dS/dt as denoting a very small displacement (dS) divided by a very short time interval (dt) without going badly wrong. [In the 20th century, mathematicians came up with theoretical support (non-standard analysis and the Transfer Principle) which can make it formally correct to consider dS/dt as the ratio between an infinitesimal displacement and an infinitesimal time].

 

[sophiecentaur]

Why 's' in all kinematics equation is displacement.

If you were to use x,y or z, it would imply motion in a straight line whereas s can be general distance in any direction or even along any path. (With suitably keeping tabs on any changes in direction and the possible effect of this on any acceleration that's involved.

 

[jbriggs444]

but I am not clear why and how S is displacement , is it because accelaration is a vector

S is displacement because we used the letter S to represent displacement. There is no deeper meaning. Displacement is a vector, yes.

 

[parshyaa]

S is ditherefore velocity t because we used the letter S to represent displacement. There is no deeper meaning. Displacement is a vector, yes.

I think generally we always treat accelaration as vector therefore velocity will also be vector and therefore S will be displacement . Do all kinematics equations are vectorial . If they are in straight line we right them without vector notation.

 

[parshyaa]

I think generally we always treat accelaration as vector therefore velocity will also be vector and therefore S will be displacement . Do all kinematics equations are vectorial . If they are in straight line we right them without vector notation.

I want to say in derivation of these equations we take a = dv/dt(which is a vector and also velocity ), therefore we get using intigration v = u + at in vector form , again takin v = ds/dt we will get s= ut + 1/2 at^2. Therefore s must be displacement

 

[jbriggs444]

I think generally we always treat accelaration as vector therefore velocity will also be vector and therefore S will be displacement .

Re-read that logic. One can conclude that "displacement is a vector", not that "S will be displacement".

 

[parshyaa]

Whats difference between these equations

• S = ut + 1/2 at^2
• And X - X' = ut + 1/2at^2.

I think here X and X' represents position of particle at time 0 and time = t respectively, and together (X-X') represents displacement and we write it as S. And in V = dx/dt , x represents position of particle at time t. I think I am right, please help me .

 

[Dr. Manoj]

Whats difference between these equations

• S = ut + 1/2 at^2
• And X - X' = ut + 1/2at^2.

I think here X and X' represents position of particle at time 0 and time = t respectively, and together (X-X') represents displacement and we write it as S. And in V = dx/dt , x represents position of particle at time t. I think I am right, please help me .

In equation 1, S represents displacement. So displacement is shortest distance between any two points. So X-X' is used to get displacement vector. Where X&X' are position vectors

 

[sophiecentaur]

In equation 1, S represents displacement. So displacement is shortest distance between any two points. So X-X' is used to get displacement vector. Where X&X' are position vectors

s has a more general definition than that. If you have a locomotive on a curved track, s can represent the displacement along the track and that will not be the distance of the point in cartesian co ordinates.

 

[Leo Michaels]

I just started learning kinematic formulas and i tend to make mistakes when it comes to applying the kinematic formulas;this is one of the questions i had trouble with:

The question provided the following variables:
Δx=110m (displacement)
Vf=34m/s (final velocity)
Vi=29m/s (initial velocity)
a=?asked by the question (acceleration)
t=(not asked or mentioned in question) (time)[not required according to circumstances of the question]

I used the the fourth kinematic formula: vf^2 = vi^2+2ad

rearranging them I get: a = ( vf^2 - vi^2 ) / 2Δx

I plug-in the variables: a = (34m/s^2) - (29m/s^2) / 2 * 110

a = 1156m/s - 841m/s /110m

I had trouble calculating so I cross checked with a calculator and i got: a = 2.9m/s^2 (approx)

but later finding out the answer was 1.4m/s^2 (approx)

a ≠ 2.9

a = 1.4

what mistake did I make in my calculations?(I even used a calculator at the end)[is using the calculator causing issues to the calculation?since it is physics?]

Please help me by posting your correct solution and how you did it and do please tell me what i did wrong.

 

[jbriggs444]

I plug-in the variables: a = (34m/s^2) - (29m/s^2) / 2 * 110

a = 1156m/s - 841m/s /110m

Did you miss something going from the one line to the next?

 

[Leo Michaels]

Did you miss something going from the one line to the next?

Ha Ha Ha Ha Ha! Thanks jbriggs444 how silly of me to miss the (2*110) part!

 

[Leo Michaels]

The answer:

a = (34m/s)^2 - (29m/s)^2 / 2*110

a= 1156m/s - 841m/s / 220

a= 1.4m/s (approx)

 

[Leo Michaels]

Did you miss something going from the one line to the next?

Cherrio jbriggs444 thanks again!

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[jbriggs444]

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[Leo Michaels]

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