How Does the Chain Rule Explain Acceleration in Terms of Distance?

[christian0710]

Hi I'm learning about The chainrule, and I understand how to apply the chain rule on various problems, but there is a problems I don't understand how works: 1) The book I'm reading writes acceleration as

a=v*(dv)/dt

And IT argues that v=ds/dt and a=dv/dt (which i understand)

So therefore by applying the chain rule we get

a=dv/ds*ds/dt =v*dv/ds.

The part i don't understand is: How did we get dv/ds by applying the chain rule? What does the derivative of Velocity with respect to distance even men? Usually it's ds/dv and not dv/ds.

 

[Orodruin]

The part i don't understand is: How did we get dv/ds by applying the chain rule?

Yes, s is a function of t and (as long as it is invertible) you can therefore write v as a function of s.

What does the derivative of Velocity with respect to distance even men?

What do derivatives usually mean?

 

[christian0710]

Yes, s is a function of t and (as long as it is invertible) you can therefore write v as a function of s.What do derivatives usually mean?

Thank you for the reply
invertible derivatives is new for me, So i assume it means the change in velocity with respect to distance but don't really see a) how it works out in practice and b) why you can just do that (there was no proof in the book):
if velocity is v=2t then s=1t^2 +Vo so a=v*dv/ds =2t(dv/ds) but how do you find dv/ds?

 

[Orodruin]

but how do you find dv/ds?

This would depend on the information you have at hand. If you have v as a function of s, you would simply perform the derivative.

 

[tommyxu3]

Sometimes the relation between $v$ and $s$ is more useful than that between $v$ and $t,$ while the latter is more often used for it illustrates $a.$

 

[GiuseppeR7]

some times you may want to realize that V is a function of S that is of course a function of t

V=V(S(T))
so taking the derivative (with respect to t) you are going to obtain:
a=(dV/dS)V

there is nothing deeper than this! The general chain rule is: having

g=g(h(x))

g'=g'(h(x))h'(x)