# Problem with velocity vector [closed]

Question:

The radius vector of a point depends on time $t$, as $\vec{r} = \vec{c}t+\dfrac{\vec{b}t^2}{2}$ where $c$ and $b$ are constant vectors. Find the magnitude of velocity.

My attempt :

$$\vec{v} =\dfrac{d\vec{r}}{dt}=\vec{c}+\vec{b}t$$

magnitude : $$\sqrt{|\vec{c}|^2 + |\vec{b}t|^2 + 2\vec{c}\cdot\vec{b}t}$$

but in solution book, the magnitude was given as :

$$\sqrt{\vec{c}\cdot\vec{c}+\vec{b}t\cdot\vec{b}t + 2\vec{c}\cdot\vec{b}t}$$

My question is which is correct?

• I don't see a difference between the two solutions. – ACuriousMind Jun 21 '15 at 17:39

The answers are actually equivalent. $|\vec{c}|^2 = \vec{c} \cdot \vec{c} = \Sigma_i x_i^2$ Where the $i$'s run over whatever number of dimensions you have. So you're both right.
• no but isnt $\vec{c} \cdot \vec{c} = |c||c|\cos\theta$ ? whis is may not equivalent to $|c||c|$ ? – Max Payne Jun 21 '15 at 17:35
• @TimKrul But what is the angle between $\vec{c}$ and it self? – fibonatic Jun 21 '15 at 17:40