Solving for distance: How do you get to v=d/t?

[dag45hol]

Forgive the basic question but my Google Fu isn't strong enough in math.

I understand that for constant velocity the function of the distance relative to the time takes the form of: d=A*t+ B

Then of course the first derivative is the velocity: d'=v=A, which is a constant (expected since we started with a constant velocity).

Q1: how do you represent d' in Leibniz's notation?
Q2: how do you go from d' to the formula v=d/t?

Thanks

 

[I like Serena]

Welcome to PF, dag45hol!

Forgive the basic question but my Google Fu isn't strong enough in math.

I understand that for constant velocity the function of the distance relative to the time takes the form of: d=A*t+ B

Then of course the first derivative is the velocity: d'=v=A, which is a constant (expected since we started with a constant velocity).

Q1: how do you represent d' in Leibniz's notation?
Q2: how do you go from d' to the formula v=d/t?

Thanks

A1: The representation of d' in Leibniz's notation would be $d' = \frac {dd} {dt}$.
Of course this looks a bit ambiguous, so the letter d for distance is usually avoided when using Leibniz's notation.
Typically one of the letters s or x is used.

A2: The formula v=d/t is not always true in your case.
It will only hold if the B in your equation is zero.
In practice B is often chosen to be zero, so the formula v=d/t does hold.

 

[HallsofIvy]

d= At+ B has a line as graph. For such a function, the derivative is a constant:
\lim_{h\to 0}\frac{(A(t+h)+ B)- (At+ B)}{h}= \lim_{h\to 0}\frac{At+ Ah+ B- At- b}{h}
= \lim_{h\to 0}\frac{Ah}{h}= A

If, with d(t)= At+ B, d(0)= B is NOT 0, v is NOT just "d/t". The average velocity, between t₀ and t₁, would be
\frac{d(t_1)- d(t_0)}{(t_1- t_0}= \frac{(At_1+ B)- (At_0+ B)}{t_1- t-0}= \frac{A(t_1- t_0)}{t_1- t_0}= A

As for Liebniz's notation, it would be dd/dt with the understanding, of course, that the second "d" is the distance function. There is no deep mathematics there but the possibility of confusion is one reason why it is more common to use "s" to represent the distance function.

 

[dag45hol]

Lets replace d(istance) with s for clarity.

I get how ds/dt=v and how it comes to be a constant (A). What I don't get is how ds/dt=v=s/t

I get the feeling that this is obvious to you guys so please have patience.Thanks

 

[Char. Limit]

The thing you have to remember is that v=s/t is not always true. v is not always simply the quotient of distance and time. However, for constant velocity, the above equation holds. With non-constant velocity, it's better to use v=ds/dt.

 

[dag45hol]

I understand that v=s/t is only good when v is a constant and that B must be zero for the function to hold.

The question is how do you go from v=ds/dt to the formula v=s/t?

Once again it seems that the answer is so obvious to you guys that you think I'm asking about other things.

ds/dt=v => fine
v=s/t => why? how do you go from ds/dt to s/t?

 

[I like Serena]

You already had d'=A=v.

So your formula d=A*t+B, can also be written as d=v*t+B.
This can be manipulated into:

d=v*t+B
d-B=v*t
(d-B)/t=v
v=(d-B)/t

With B is 0, this becomes v=d/t.

 

[dag45hol]

Thanks, I knew it had to be something obvious =)