# Using calculus to find velocity

1. May 30, 2014

### albertrichardf

Hi all,
Supposing that s' is the new position, s the old one, t the time started, t' the new time, Δt the time difference, (approaches zero) and v the velocity, or the derivative of the position with respect to time: s' = s + vt. So far so good, I'll work out the derivative first:
ds/dt = vt + vΔt - vt/vΔt
= v.
That shows that the derivative of ds/dt is v. However, most calculus equations are given as s = ƒ(t) which can be something like s = 4t^2 + 2t + 20. How is this possible? I know that the equation of a straight line can be written as : y = mx + c which in this case would be s' = vt + s. So how is it possible, without knowing the velocity, to find an equation such as the s = ƒ(t)? one here? Besides doesn't this equation hold only for a straight line? So would it apply throughout an entire graph if the velocity, that is the slope changes? And if so, why?
Any help would be greatly appreciated, thanks

2. May 30, 2014

### HallsofIvy

You appear to be assuming that v is constant.

If you are "given" s= 4t^2+ 2t+ 20 then v= ds/dt= 8t+ 2.

If you do not "know the velocity" how you find s=f(t) depends upon what you are given!

I really don't know what you are asking! "One" what?

What equation are you referring to? If you have a graph of s= s(t), then ds/dt is the slope of the tangent to the graph at each s. If the graph itself is a line then that "tangent" is the line itself.

3. May 30, 2014

### albertrichardf

That shows that the derivative of ds/dt is v. However, most calculus equations are given as s = ƒ(t) which can be something like s = 4t^2 + 2t + 20. How is this possible?

This was actually meant as an example of the sort of relationship given between the displacement or distance and the time in most physics calculus questions. The question was actually how is this derived from the derivation shown here:

ds/dt = vt + vΔt - vt/vΔt
= v.

I'm pretty new to calculus so I'm not sure if this derivation assumes the velocity to be constant. If so, would it be possible to show how the equation should be f velocity is not constant. I would appreciate if that was done in the same format as the one given here, that is an equation formed rather than the use of laws such as the law saying that d(xa) is equal to axa-1. Thanks

one here?

I actually meant the function that was given in the question. For clarity and convenience I will quote it here:

However, most calculus equations are given as s = ƒ(t) which can be something like s = 4t^2 + 2t + 20

So the one here refers to the s = ƒ(t) shown in the quote above.

Besides doesn't this equation hold only for a straight line? So would it apply throughout an entire graph if the velocity, that is the slope changes? And if so, why?
Any help would be greatly appreciated, thanks

What equation are you referring to? If you have a graph of s= s(t), then ds/dt is the slope of the tangent to the graph at each s. If the graph itself is a line then that "tangent" is the line itself.

the equation I meant was the following:
y = mx + c where y is the y-coordinate, m the gradient, x the x-coordinate and c the y-intercept.
This applies to a straight line only. So if ds/dt is the gradient of a line tangent to a slope, that means that ds/dt will change with change in velocity since there will only be a slope if velocity s changing, which would be defined as acceleration. Therefore there is no constant relationship between the distance and the time if velocity changes. Please correct me if I am wrong, or validate me if I'm right. Any help would be appreciated, thanks.

4. May 30, 2014

### HallsofIvy

Perhaps it was a misprint but it is NOT "the derivative of ds/dt" that is v. The derivative of v, which is dv/dt, is v.
I don't know what you are asking. Pretty much any function can be a distance function. If, in this case, s= 4t^2+ 2t+ 20 then ds/dt= v= 8t+ 2 and the acceleration is dv/dt= 8.

I really don't know where you got "vt+ vΔt - vt/vΔt". v is the speed at instant t. What does Δt mean here?

Yes, if velocity is changing, then you cannot write s as vt+ b because v is NOT constant.
A standard example is an object falling under gravity, neglecting air friction. The acceleration is a= dv/dt= -g so v=-gt+ v0 is a linear function. Taking the anti-derivative again, s= -(g/2)t^2+ v0t+ s0.

5. May 30, 2014

### albertrichardf

I actually meant that the instantaneous velocity is equal to ds/dt as dt approaches 0.

This was actually meant as an example of the sort of relationship given between the displacement or distance and the time in most physics calculus questions. The question was actually how is this derived from the derivation shown here:

ds/dt = (vt + vΔt - vt)/vΔt
= v.
I really don't know where you got "vt+ vΔt - vt/vΔt". v is the speed at instant t. What does Δt mean here?

To derive this, here is what I did:

s' the new value of s = s + Δs, where Δs is the change in position
s' = ƒ(t + Δt) where saying that v is constant, ƒ(t + Δt) = vt = vΔt + s, where Δt represents a change in time
and therefore s = ƒ(t) which = vt + s.

since m, the gradient = Δs/Δt and Δs = s' - s:

m =[ ƒ(t + Δt) - ƒ(t) ] / Δt or:

[ vt + vΔt + s - (vt + s) ] / Δt
= [ vt + vΔt + s - vt - s ] / Δt
And simplify:
= vΔt/ Δt (vt - vt and s - s cancel out)
= v
So m, the gradient = v
Δt is actually a short time interval, which as it approaches zero, that is the limit of t is 0, gives the instantaneous velocity. Δt is actually meant to be seen as dt.

I think the original equation I gave earlier s = 4t2 + 2t + 20 actually originates from the equation s' = 1/2 at2 + vt + s where here 4 is 1/2 the acceleration, 2 the velocity and 20 the initial position.

So based on this, I suppose that the acceleration must be known when finding this equation, as well as the initial velocity and the initial position. However by use of calculus, the acceleration can be found again from the distance equation as well as the instantaneous velocity.
Thanks, that really clarified things. This seems a lot less confusing now

6. May 31, 2014

### HallsofIvy

Badly stated. "instantaneous velocity" is NOT "ds/dt" as dt approaches 0. It is ds/dt at a given value of t. I presume you were thinking of the formula for ds/dt which is $\lim_{\Delta t\to 0} \frac{f(t+\Delta t)- f(t)}{\Delta t}$
But it is not "dt" that is "approaching 0"..

This is not true. You can write "$v\Delta t+ s$" if v is constant but then "vt" is not equal to that.

Which is neither of the things you wrote.

No, the gradient is NOT $$\Delta s/\Delta t$$, it is the limit of that.

Again, all of that is true only as long as v is constant.

No, $\Delta t$ is not "meant to be seen" as dt. You are misunderstanding the notation.

7. Jun 1, 2014

### albertrichardf

I actually meant that the instantaneous velocity is equal to ds/dt as dt approaches 0.
Badly stated. "instantaneous velocity" is NOT "ds/dt" as dt approaches 0. It is ds/dt at a given value of t. I presume you were thinking of the formula for ds/dt which is limΔt→0f(t+Δt)−f(t)Δt
But it is not "dt" that is "approaching 0"..

Yeah, thanks for the correction. As I said, my calculus is quite rudimentary, so I actually thought it meant the same thing.

(t + Δt) where saying that v is constant, ƒ(t + Δt) = vt = vΔt + s

It actually is a typo. The corrected equation:

ƒ(t + Δt) = vt + vΔt + s

That does make a lot of things to recheck for me. But actually my question actually meant this:
Without knowing, the acceleration, the initial velocity nor the initial displacement, how is it possible to express s as a function of t?.

That basically is it. Since s expressed as t depends on the acceleration, the initial velocity and the initial displacement, how can an equation be formed without the knowledge of those factors?

8. Jun 1, 2014

### Shinaolord

You integrate twice.

If acceleration $a=a(t)$ then $v(t)$ is defined as
$v(t)=\int a(t) dt =a(t)*t+v$
Where v is the velocity at t=0 (the initial velocity)
It doesn't matter what the function a(t), is. Velocity will always be defined as $\int a(t) dt$ called the antiderivative.

I'm on my phone, I will finish this post in a moment.

If we want a function for displacement $s(t)$, we just take the antiderivative of $v(t)$

$s(t)= \int v(t) = \frac{a(t)}{2} ^2 + v(t)*t + s$ where s is initial displacement.

Last edited: Jun 1, 2014
9. Jun 1, 2014

### Shinaolord

As a correction to my previous post, velocity is not always defined as $\int_{a}^{b} a(t) dt$, we can also define is as the derivative of s(t)
$\frac{d}{dt} s(t) = v(t)$
Likewise with $v(t)$ for $a$.
Please note this is exactly what I did in my original post, but reversed .
differentiation and anti-differentiation are inverses of each other.
I'm not sure how to do limits in LateX or I would give you the limit definition.

Last edited: Jun 1, 2014
10. Jun 2, 2014

### albertrichardf

It does help a little. However it does't answer it entirely. Here is an example:
An object's position is described by the following polynomial for 0 to 10 s.
s = t^3 − 15t^2 + 54t.

My question is, how was this found? That is, how to describe s as a function of time? I know that there is no specific function for all s-t relationships, but I mean for a given set of values s, and the corresponding t, how do we express s as a function of t?

For example, suppose we are given this:
t = 0, s = 0
t = 1 s = 1
t = 2 s = 4
t = 3 s = 9
t = 4 s = 16

Here it is obvious that s = t^2, but what if the relationship between s and t is less obvious, such as in the example in pink? Is there a way of finding the relationship between the two or can this be done only through trial and error?

11. Jun 2, 2014

### Shinaolord

Last edited: Jun 2, 2014
12. Jun 2, 2014

### Staff: Mentor

This integration is not correct unless a(t) is constant (i.e., independent of t).

This integration is also incorrect.

Chet

13. Jun 3, 2014

### Shinaolord

Ahh! What was I thinking ? You're absolutely correct that it has to be time independent; a constant.
I guess I should think these things through a lot more before I post my attempts at helping others.
My apologies to the OP.