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 solutionPhysics

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