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Question

The position of a particle is given by  $$\vec{r}=\left( 3t\hat{i}-{{t}^{2}}\hat{j}+4\hat{k} \right)$$  , what will be the velocity and acceleration?


Solution

Given that,

Position vector $$\vec{r}=\left( 3t\hat{i}-{{t}^{2}}\hat{j}+4\hat{k} \right)$$

Now, the velocity is

  $$ \vec{v}=\dfrac{d\vec{r}}{dt}=\dfrac{d}{dt}\left( 3t\hat{i}-{{t}^{2}}\hat{j}+4\hat{k} \right) $$

 $$ v=\left( 3\hat{i}-2t\hat{j} \right)\,m/s $$

Now, the acceleration is

  $$ \vec{a}=\dfrac{d\vec{v}}{dt} $$

 $$ \vec{a}=\dfrac{d}{dt}\left( 3\hat{i}-2t\hat{j} \right) $$

 $$ \vec{a}=\left( -2\hat{j} \right)\,m/{{s}^{2}} $$

Hence, this is the required solution


Physics

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