# Homework Help: Mechanics (Components of Acceleration)

1. Apr 20, 2012

### roam

1. The problem statement, all variables and given/known data

The following is a solved problem:

http://img600.imageshack.us/img600/6503/questiono.jpg [Broken]

I can't understand how they worked out these two components.

2. Relevant equations

$\vec{a}(t) = \frac{dv}{dt}\vec{T} + v \frac{d \vec{T}}{dt}$

Where

$\vec{T}(t) = \frac{d \vec{r}(t)/dt}{||d \vec{r}(t)/dt||} = \frac{\vec{v}(t)}{v(t)}$

3. The attempt at a solution

In the equation I posted above, I believe the first term is the tangential component of acceleration and the second one is the normal component. I want first to find the tangential component:

$\vec{v}(t) = d(t^2 \vec{i} + 3t+2 \vec{j})/dt = 2t+3$

$|| \vec{v}(t) || = \sqrt{4t^2+9}$

So substituting in

$\frac{dv}{dt}\vec{T} = \frac{d (\vec{r}(t)/dt)}{dt} . \frac{d \vec{r}(t)}{dt}.\frac{1}{||\vec{r}(t)||}$

$2.(2t+3).\frac{1}{\sqrt{4t^2+9}}$

$= \frac{4t+6}{\sqrt{4t^2+9}}$

But this is wrong. The correct answer must be

$\frac{4t}{\sqrt{4t^2+9}}$

What did I do wrong? How can I get to the correct answer? Any guidance is greatly appreciated.

Last edited by a moderator: May 5, 2017
2. Apr 21, 2012

### Quinzio

You have speed: $\vec v= \frac{dr}{dt}=(2t,3)$
and acceleration: $\vec a= \frac{dv}{dt}=(2,0)$

then you've got to find the projection of $\vec a$ onto $\vec v$:
$$\vec a_t = \frac{(\vec a \cdot \vec v)}{||\vec v||^2}\vec v= \frac{1}{4t^2+9}(8t^2,12t)$$
The normal component of acceleration is then:
$\vec a_n= \vec a - \vec a_t$

The given answers seem to be wrong.

Last edited by a moderator: May 5, 2017