Question 184 of 308

The position vector of a moving body at any instant of time is given as $\vec{r}=\left(5 t^2 \hat{i}-5 t \hat{j}\right) \mathrm{m}$. The magnitude and direction of velocity at $t=2 \mathrm{~s}$ is,

A$5 \sqrt{15} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with - ve $Y$ axisB$5 \sqrt{15} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with + ve X axisC$5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with + ve $X$ axisD$5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with - ve $Y$ axis

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Answer: (D) $5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with - ve $Y$ axis

$\begin{aligned} & \overrightarrow{\mathrm{r}}=5 t^2 \hat{\mathrm{i}}-5 \mathrm{t}_{\mathrm{j}} \\ & \overrightarrow{\mathrm{v}}=10 \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{v}}=20 \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \quad \text { at } \mathrm{t}=2 \mathrm{sec}\end{aligned}$



$\begin{aligned} & \tan \theta=\frac{20}{5}=4 \\ & \theta=\tan ^{-1} 4 \\ & \text { From -veY-axis }\end{aligned}$