SATHEE: Kinematics Question 150

Kinematics Question 150

Question: The position vector of a particle is determined by the expression $\vec{r} = 3 t^{2} \hat{i} + 4 t^{2} \hat{j} + 7 \hat{k}$ The distance traversed in first 10 sec is [DPMT 2002]

Options:

A) 500 m

B) 300 m

C) 150 m

D) 100 m

Show Answer

Answer:

Correct Answer: A

Solution:

$\vec{r} = 3 t^{2} \hat{i} + 4 t^{2} \hat{j} + 7 \hat{k}$ at $t = 0$ , ${\vec{r}}_{1} = 7 \hat{k}$ at $t = 10 \sec$ , ${\vec{r}}_{2} = 300 \hat{i} + 400 \hat{j} + 7 \hat{k}$ , $\vec{Δ r} = {\vec{r}}_{2} - {\vec{r}}_{1} = 300 \hat{i} + 400 \hat{j}$

$| \vec{Δ r} | = | {\vec{r}}_{2} - {\vec{r}}_{1} | = \sqrt{{(300)}^{2} + {(400)}^{2}} = 500 m$

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Kinematics Question 150

Question: The position vector of a particle is determined by the expression r→=3t2i^+4t2j^+7k^ The distance traversed in first 10 sec is [DPMT 2002]

Options:

Answer:

Solution:

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Question: The position vector of a particle is determined by the expression $\vec{r} = 3 t^{2} \hat{i} + 4 t^{2} \hat{j} + 7 \hat{k}$ The distance traversed in first 10 sec is [DPMT 2002]