Kinematics Question 592
Question: The position of particle is given by $ \vec{r}=2t^{2}\widehat{i}+3t\widehat{j}+4\widehat{k}, $ where $ t $ it’s in second and the coefficients have proper unit for $ \vec{r} $ to be in meter. The $ \vec{a}(t) $ of the particle at $ t=1s $ it’s
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
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \vec{r}=2{{t^2}}\hat{i}+3t\hat{j}+4\hat{k} $
$ \therefore {\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} $
$ {\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} $
