Vector analysis question on acceleration

[jaejoon89]

Homework Statement

A moving particle reaches its max. speed at the instant t = 3. (Before and after 3, its speed is less.) It follows that the particle's acceleration is 0 at the instant t = 3... Show that this is FALSE

Homework Equations

v = dR/dt
a = d^2 R / dt^2

The Attempt at a Solution

How do I show this is false? The derivative of velocity is acceleration so I would think it's true and is indeed 0. This is on a chapter for vector analysis on acceleration and curvature.

 

[gabbagabbahey]

If the particle's acceleration were 0 at t=3, why would it slow down?

 

[jaejoon89]

But the acceleration has to be 0 because at this point the speed is maximum... so on one side there should be negative acceleration and on the other side positive.

 

[gabbagabbahey]

No, if the acceleration at t=3 were zero, then if you measured the speed of the particle a very short time later it would be unchanged.

It's true that \frac{d}{dt} ||\vec{v}||=0 at t=3, but that doesn't necessarily mean ||\vec{a}||=0 at t=3.

This rests on the fact that \vec{v} is a vector, it has both magnitude and direction and just because its magnitude isn't changing doesn't mean it's direction can't be changing. If it's direction is changing, then \vec{a}=\frac{d\vec{v}}{dt}\neq0

There is an expression for \vec{a} in terms of its tangential and normal components that you should know (it involves curvature), use that to prove that the acceleration is non-zero at t=3!