Q. The position vector of a particle is determined by the expression $ \vec{r}=3t^{2}\hat{i}+4t^{2}\hat{j}+7\hat{k}$. The distance traversed in first $10\, s$ is :
Delhi UMET/DPMTDelhi UMET/DPMT 2002
Solution:
The position vector is the vector from the origin of the coordinate system (0, 0, 0) to the point $P,(x, y, z)$.
The vector change in position, from origin gives the distance traversed.
Given, $ \vec{r}=3t^{2}\hat{i}+4t^{2}\hat{j}+7\hat{k}$
At $t = 0$, we get $\vec{r}_{0}=7\hat{k}$
At $t = 10\, s$, we get $\vec{r}=3\times(10)^{2}\hat{i}+4\,(10)^{2}\hat{j}+7\hat{k}$
$=300\,\hat{i}+400\hat{j}+7\hat{k}$
Distance travelled, $\vec{s}=\vec{r}-\vec{r}_{0}$
$=300\hat{i}+400\hat{j} +7\hat{k}-7\hat{k}$
$=300\hat{i}+400\hat{j}$
Hence, $\left| \vec{s} \right|=\left| \vec{r}-\vec{r}_{0} \right|$
$=\sqrt{{(300)}^{2}}+{{(400)}^{2}}$
$=500\,m$
