JEE MAIN - Physics (2025 - 24th January Evening Shift - No. 17)
The position vector of a moving body at any instant of time is given as $\overrightarrow{\mathrm{r}}=\left(5 \mathrm{t}^2 \hat{i}-5 \mathrm{t} \hat{j}\right) \mathrm{m}$. The magnitude and direction of velocity at $t=2 s$ is,
$5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with - ve Y axis
$5 \sqrt{15} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with + ve $X$ axis
$5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with + ve $X$ axis
$5 \sqrt{15} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with $-$ ve $Y$ axis
Explanation
$$\begin{aligned} & \overrightarrow{\mathrm{r}}=5 \mathrm{t}^2 \hat{\mathrm{i}}-5 \hat{\mathrm{t}} \\ & \overrightarrow{\mathrm{v}}=10 \mathrm{t} \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{v}}=20 \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \quad \text { at } \mathrm{t}=2 \mathrm{sec} \end{aligned}$$
_24th_January_Evening_Shift_en_17_1.png)
$$\begin{aligned} & \tan \theta=\frac{20}{5}=4 \\ & \theta=\tan ^{-1} 4 \\ & \text { From-veY-axis } \end{aligned}$$
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