Physics Two Dimensional Motion Question 140
Question: The position of particle is given by $ \vec{r}=2,t^{2}\widehat{i}+3,t\widehat{j}+4\widehat{k}, $ where $ t $ is in second and the coefficients have proper units for $ \vec{r} $ to be in meter. The $ \vec{a},(t) $ of the particle at $ t=1s, $ is
Options:
A) $ \text{4},\text{m }{{\text{s}}^{-2}} $ along y-direction B) $ \text{3},\text{m }{{\text{s}}^{-2}} $ along x-direction C) $ \text{4 m }{{\text{s}}^{-2}} $ along x-direction D) $ \text{2 m }{{\text{s}}^{-2}} $ along z-direction
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Answer:
Correct Answer: C
Solution:
[c] $ \vec{r}=2{{t^{2}}\hat{i}+3t\hat{j}+4\hat{k} $
$ \
Therefore \text{\vec{v}=}\frac{d\vec{r}}{dt}=\frac{d}{dt}=( 2{{t^{2}}\hat{i}+3t\hat{j}+4\hat{k} )=4\text{t}\hat{i}+3\hat{j} $
$ \text{\vec{a}=}\frac{d\vec{v}}{dt}=\frac{d}{dt}( 4\text{t}\hat{i}+3\hat{j} )=4\hat{i} $
$ \
Therefore \vec{a}=4m{{s}^{-2}}\text{ along x-direction} $
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