2

In one dimension, the acceleration of a particle can be written as::

a = dv/dt = vdv/dx

Does this equation imply that, if v=0 , then a = 0 ?

But I can think of several situations when object is at instantaneous rest and a is not 0 ? What is going on here ?

3

Considering the equation a = vdv/dx, the correct thing to say would be if v = 0 and dv/dx if finite, then a would be zero

As if v = 0 and dv/dx if infinite, then it would be ( 0 x infinite) term, which is not actually 0, so wont't make our 'a' zero

Now, let us see what happensConsider the case of an object thrown upwards, where a = -g

s = ut - (1/2)gt^{2}

v = u - gt

We will get:

v^{2} = u^{2} - 2gs

Taking square root and differentiating it:

dv/dx = +-g / (sqrt ( u^{2} - 2as )

So it concludes that, in a = vdv/dx, if v = 0 then dv/dx is infinite, which does not make a = 0.

0

When you apply a=dv/dt you also have to think about just before and just after the instant at which you are trying to find the

acceleration. If velocity is zero at the instant and is also 0 just before and just after the instant then the acceleration is 0. As for example an object is on rest. In this case acceleration is clearly 0.

But let us throw a particle upward. Now at the highest point of motion its velocity is 0 but its acceleration is g. That's because just before or just after that instant the velocity is not 0 and so its acceleration is nonzero.

I hope it helps.