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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?

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

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  • $\begingroup$ no but isnt $\vec{c} \cdot \vec{c} = |c||c|\cos\theta$ ? whis is may not equivalent to $|c||c|$ ? $\endgroup$ – Max Payne Jun 21 '15 at 17:35
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    $\begingroup$ @TimKrul But what is the angle between $\vec{c}$ and it self? $\endgroup$ – fibonatic Jun 21 '15 at 17:40
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The answers are exactly equivalent, the dot product between two vectors produces the product between magnitude of the vectors and the cosine of the angle between the vectors...

In this case the angle between two c vectors will be 0 and the angle between two (bt) vectors will also be 0, hence the dot product will be equal to the square of the magnitude of the vectors.

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