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…

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,
  1. $5 \sqrt{15} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with - ve $Y$ axis
  2. $5 \sqrt{15} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with + ve X axis
  3. $5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with + ve $X$ axis
  4. $5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with - ve $Y$ axis

Solution

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

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

Asked in: JEE Main 2025 (24 Jan Shift 2)

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