Question

The radius vector of a point depends on time t, as :
$\bullet \: \sf r = ct + \dfrac{b {t}^{2} }{2}$
Where c and b are constant vectors. Find the modulus of Velocity and a Acceleration at any time t.​

Answer 1

Answer:

It is the correct answer.

Explanation:

Hope this attachment helps you.

Answer 2

$\large{ \bf{ \green { \mathfrak{ \dag{ \underline{ \underline{ \sf \: Given}}}}}}}$

• $\sf r = ct + \dfrac{b {t}^{2} }{2}$

$\sf \: V = \frac{dr}{dt} = > C + 2bt \\ \sf \: acceleration = \frac{dv}{dt} = > 0 + 2b \\ : \implies \boxed{ |a| = 2b}$

$\sf \: V = C + tb \sf \red{(given)}$

$\sf \: = \sqrt{C {}^{2} + (2tb) {}^{2} + 2 \times C \times 2bt \times \cos \theta }$

$\boxed{\boxed{\sf\overrightarrow{\boldsymbol{C}}.\sf\overrightarrow{\boldsymbol{C}} = C {}^{2} } \: \boxed{\sf\overrightarrow{\boldsymbol{b}}.\sf\overrightarrow{\boldsymbol{b }} = b {}^{2} } \: \boxed{\sf\overrightarrow{\boldsymbol{C}}.2\sf\overrightarrow{\boldsymbol{tb}} = 2t \times cb \cos \theta}} \:$

We can write it like this ;

$\rightarrow \: \sf\sqrt{C.C + 4t {}^{2} \: b.b + 2 \times \: 2t\sf\overrightarrow{\boldsymbol{C}}.\sf\overrightarrow{\boldsymbol{b}}} \sf \\ \implies \boxed{ \boxed{ \sf\green{ \sqrt{ {\sf \: C.C + 4t {}^{2} b.b + 4t\sf\overrightarrow{\boldsymbol{C}}.\sf\overrightarrow{\boldsymbol{b}} }}}}}$

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$\sf\blue{hope \: this \: helps \: you!! \: }$