Acceleration as derivate of velocity

[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?

 

[ZapperZ]

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

This, btw, is a mathematics question.

Zz.

 

[mentalguy]

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

This, btw, is a mathematics question.

Zz.

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?

 

[HallsofIvy]

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).

 

[ZapperZ]

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?

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.

 

[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 ?

 

[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.

 

[ZapperZ]

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 ?

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.