Question

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 $10s$ is :

• A. 500 m
• B. 300 m
• C. 150 m
• D. 100 m

Answer 1

Answer: (A)

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 $t=10s$, 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, $\left|\vec{s}\right|=\left|\vec{r}-{\vec{r}}_0\right|$
$=\sqrt{{(300)}^2}+{(400)}^2$
$=500m$