# Acceleration as derivate of velocity

1. Nov 18, 2013

### mentalguy

I have problem understanding the following derivation:

a = dv/dt
v= ds/dt
(i get this part)

but then,

a = (dv/ds)(ds/dt)

I can't understand the above equation? Please can you tell me how 'a' gets that value?

2. Nov 18, 2013

### ZapperZ

Staff Emeritus
Have you done a section on calculus called the chain rule?

This, btw, is a mathematics question.

Zz.

3. Nov 18, 2013

### mentalguy

yes i know the chain rule

if y=f(g(x)) then dy/dx = df/dg * dg/dx

but confused that how is that applied here?

4. Nov 18, 2013

### HallsofIvy

Staff Emeritus
s(t) (distance moved) and v(t) (speed) are are both functions of t. You can then think of v as a function of t. (For each s, find the corresponding t, then find v(t) for that t.) So dv/ds is defined and by the chain rule, a= dv/dt= (dv/ds)(ds/dt).

5. Nov 18, 2013

### ZapperZ

Staff Emeritus
I don't understand the problem here.

To be able to write that chain rule, it means that you have a function v(s). Thus, since a = dv/dt, use the chain rule to write

a= dv/ds * ds/dt.

Zz.

6. Nov 18, 2013

### mentalguy

What i am saying is that there is a function that v(s(t)) [v is a function of s which is a function of t]

So in that case, when chain rule is applied i get a= dv/ds * ds/t

Am i right in my thinking ?

Last edited: Nov 18, 2013
7. Nov 18, 2013

### rcgldr

If s is a function of t, then v(t) = ds/dt, and a = dv/dt = d(ds/dt)/dt = d(ds)/(dt)^2.

In the case where a is a function of s, then multiplying dv/dt by ds/ds can be used to solve the problem (if the produced integrals are solvable). The first step is to produce an equation that can be integrated (don't forget to include the constant of integration after doing the integration):

a(s) = (dv/dt)(ds/ds) = (ds/dt)(dv/ds) = v dv/ds
a(s) ds = v dv

v dv = a(s) ds

1/2 v^2 = integral(a(s) ds) + constant
v = ± sqrt(2 (integral(a(s) ds) + constant))

Assuming (a(s) ds) can be integrated, the next integration step:

define v(s) = ± sqrt(2 (integral(a(s) ds) + constant))
v(s) = ds/dt
dt = ds/v(s)
t = integral(ds/v(s)) + constant)

Even if (ds / v(s)) can be integrated, it may not be possible to invert the equation to produce s as a function of t.

Last edited: Nov 18, 2013
8. Nov 18, 2013

### ZapperZ

Staff Emeritus
I'm utterly puzzled here. Isn't that what I actually wrote (minus the "/t" in your post, which I assumed is a typo of "/dt")?

Zz.