The position vector of a particle is given by | vedclass

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(A) The position vector is given by $\vec{r}(t) = 3t^2 \hat{i} + 4t^2 \hat{j} + 7 \hat{k}$.At $t = 0 \, s$,the initial position is $\vec{r}_1 = 3(0)^2

Vedclass · Accepted Answer

$500$

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(A) The position vector is given by $\vec{r}(t) = 3t^2 \hat{i} + 4t^2 \hat{j} + 7 \hat{k}$.At $t = 0 \, s$,the initial position is $\vec{r}_1 = 3(0)^2 \hat{i} + 4(0)^2 \hat{j} + 7 \hat{k} = 7 \hat{k}$.At $t = 10 \, s$,the final position is $\vec{r}_2 = 3(10)^2 \hat{i} + 4(10)^2 \hat{j} + 7 \hat{k} = 300 \hat{i} + 400 \hat{j} + 7 \hat{k}$.The displacement vector is $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1 = (300 \hat{i} + 400 \hat{j} + 7 \hat{k}) - (7 \hat{k}) = 300 \hat{i} + 400 \hat{j}$.Since the particle moves in a straight line (the ratio of components $x$ and $y$ is constant),the distance traversed is equal to the magnitude of the displacement.Distance $= |\Delta \vec{r}| = \sqrt{(300)^2 + (400)^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, m$.

The position vector of a particle is given by the expression $\vec{r} = 3t^2 \hat{i} + 4t^2 \hat{j} + 7 \hat{k}$. The distance traversed in the first $10 \, s$ is ......... $m$.

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