# Why does acceleration = $v\frac{dv}{dx}$

If we define $x$ = displacement, $v$ = velocity and $a$ = acceleration then I am used to the ideas that $a= \frac{dv}{dt} = \frac{d^2x}{dt^2}$

However I also understand $a=v \frac{dv}{dx}$. Can someone explain to me why this is?

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## 1 Answer

Notice: $\frac{dv}{dt}=\frac{dx}{dt}\frac{dv}{dx}$ by Chain rule.

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As often with 1-variable chain rule, there are nice simple examples: if we’re travelling at 30 m/s (i.e. v = 30 m/s), and getting 2 m/s faster for each metre we travel (i.e. $\frac{dv}{dx}$ = 2 (m/s)/m), then how fast are we accelerating? Well, in a second, we’ll travel approx. 30m, gaining approx. 2 m/s for each of those metres travelled; so we’ll have gained about 30 × 2 m/s in total. So, acceleration is about 30 × 2 = 60 m/s. – Peter LeFanu Lumsdaine Feb 9 '11 at 18:30