# How Calculus works?

Dear Sirs,

I am wondering what is the difference between v=s/t & dv = ds/dt where v-velocity,s-displacement,t-time.

Consider Gallelio's equations->
v=u+at--(equation-1) &
s=ut+1/2at^2--(equation-2)
where u-initial velocity & a-acceleration

My doubt is on following point;

If we do,v = s/t in equation-2,we get->v=u+1/2at not v=u+at
but if we do,dv=ds/dt in equation-2,we get=>v=u+at ,exactly the equation we want.

why only differentiation give the result,not just mere divison?????

arildno
Homework Helper
Gold Member
Dearly Missed
v=s/t is the AVERAGE velocity over the total time INTERVAL "t".
Thus, it does NOT give the velocity at the INSTANT "t"

I am wondering what is the difference between v=s/t & dv = ds/dt where v-velocity,s-displacement,t-time.

Hi !
dv = ds/dt has no meaning at all !
dv is an infinitesimal value (a very small variation of v). Do not confuse it with v.
ds is a small displacement during dt a small variation of time.
So, ds/dt is the speed at time t, wich is not infinitesimal, hense not equal to dv.
Do not confuse the average speed s/t with the instantaneous speed ds/dt. Use two different symbols, not v for both.
dt is the small variation of speed during a small variation of time. So, dv/dt is the acceleration.

Last edited:
dv/dt is at an instantaneous time... Not division.

Mark44
Mentor
dv/dt is at an instantaneous time... Not division.
It's the acceleration at an arbitrary time.

UltrafastPED
equation $v=u+at$ is merely rearranged from the average acceleration formula $\displaystyle\frac{v-u}{t}=a$.
In calculus we get the instantaneous versions of these, $\displaystyle\frac{dv}{dt} = a$, and $v(t)=\displaystyle\int a(t) dt$