Kinematics Question 150
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 10 sec is [DPMT 2002]
Options:
A) 500 m
B) 300 m
C) 150 m
D) 100 m
Correct Answer: A $ \vec{r}=3t^{2}\hat{i}+4t^{2}\hat{j}+7\hat{k} $ at $ t=0 $ , $ {{\vec{r}}_1}=7\hat{k} $ at $ t=10\sec $ , $ {{\vec{r}}_2}=300\hat{i}+400\hat{j}+7\hat{k} $ , $ \overrightarrow{\Delta r}={{\vec{r}}_2}-{{\vec{r}}_1}=300\hat{i}+400\hat{j} $ $ |\overrightarrow{\Delta r}|=|{{\vec{r}}_2}-{{\vec{r}}_1}|=\sqrt{{{(300)}^{2}}+{{(400)}^{2}}}=500m $
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