A particle is moving in x y- plane as \overrightarrow{x}=(4 t+t^{2}) \stackrel{\^}{i}, \overrightarrow{y}=(2 t+\frac{t^{2}}{2}) \stackrel{\^}{j} where \overrightarrow{x} \& \overrightarrow{y} are displacements measured along {x} and {y} axes respectively in meters and {t} in seconds. What is the velocity of the particle?

Correct Answer is : →v=(4+2t)∧i+(2+t)∧j

\overrightarrow{v}=(4+t) \hat{i}+(2+t) \hat{j}

\overrightarrow{v}=(4+2 t) \hat{i}+(2+t) \hat{j}

\overrightarrow{v}=(4+2 t) \hat{i}+(2+\frac{t}{2}) \hat{j}

\overrightarrow{v}=(4+t) \hat{i}+(2+\frac{t}{2}) \hat{j}

TS EAMCET 20-July-2022 Shift 2 Solved Paper

Section: Physics

Question : 85 of 160

Marks: +1, -0

A particle is moving in xy− plane as

→

=(4t+t2)

→

=(2t+‌

)

where

→

are displacements measured along x and y axes respectively in meters and t in seconds. What is the velocity of the particle?

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