Question

Hey mates , anyone plz answer this question??​

Answer 1

▶ ANSWER-1) :

✴ Given :

▪ A particle is moving in three dimensions. Its position vector is provided as

$\underline{\boxed{\bf{\pink{\vec{r}=6\hat{i}+(3+4t)\hat{j}-(3+2t-t^2)\hat{k}}}}}$

✴ Calculation :

$\implies\sf\:\vec{v}=\dfrac{d\vec{r}}{dt}\\ \\ \implies\sf\:\vec{v}=\dfrac{d(6\hat{i}+(3+4t)\hat{j}-(3+2t-t^2)\hat{k})}{dt}\\ \\ \implies\sf\:\vec{v}=4\hat{j}-(2-2t)\hat{k}\\ \\ \implies\boxed{\bf{\red{\vec{v}=4\hat{j}+4\hat{k}}}}\\ \\ \implies\sf\:|\vec{v}|=\sqrt{(4)^2+(4)^2}\\ \\ \implies\boxed{\bf{\blue{v=4\sqrt{2}\:mps}}}\\ \\ \implies\sf\:\vec{a}=\dfrac{d\vec{v}}{t}\\ \\ \implies\sf\:\vec{a}=\dfrac{d(4\hat{j}-(2-2t)\hat{k})}{dt}\\ \\ \implies\boxed{\bf{\green{\vec{a}=2\hat{k}}}}\\ \\ \implies\sf\:|\vec{a}|=\sqrt{(2)^2}\\ \\ \implies\boxed{\bf{\orange{a=2\:ms^{-2}}}}$

▶ ANSWER-2) :

✴ Given :

▪ A particle is moving along z-axis, and its position vector is provided as

$\underline{\boxed{\bf{\purple{\vec{r}=(t^2-2t-3)\hat{k}}}}}$

✴ Calculation :

$\implies\sf\:v=\dfrac{dr}{dt}\\ \\ \implies\sf\:v=2t-2\\ \\ \dag\sf\:for\:rest\:position\:v=0\\ \\ \implies\sf\:2t-2=0\\ \\ \implies\boxed{\bf{\gray{t=1s}}}$