If the position vector of a particle is

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Question

$\overrightarrow{\mathrm{V}}=\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}=\sin {\mathrm{t} \hat{\mathrm{i}}}+\cos \hat{\mathrm{t} {\math

Vedclass · Accepted Answer

$1$

Vedclass · Answer

$\overrightarrow{\mathrm{V}}=\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}=\sin {\mathrm{t} \hat{\mathrm{i}}}+\cos \hat{\mathrm{t} {\mathrm{j}}}-18 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \overrightarrow{\mathrm{V}}}{\mathrm{dt}}=\cos t \hat{\mathrm{i}}-\sin t \hat{\mathrm{j}}$

$|\overrightarrow{\mathrm{a}}|=\sqrt{\cos ^{2} t+(-\sin t)^{2}}=1$

If the position vector of a particle is

$\vec r = - \cos \,t\hat i + \sin \,t\hat j - 18\,t\hat k$

then what is the magnitude of its acceleration ?

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